A.3 Chapter 3

(i) Let $a$, $r\in\mathbb{R}$ with $r\neq 1$ and, for $n\in\mathbb{N}$, consider the partial sum $s_{n}:=\sum_{k=1}^{n}ar^{k}$ of the series $\sum_{k=1}^{\infty}ar^{k}$. It follows that

Relabelling the index of summation,

Since the middle two terms cancel, we have $s_{n}-rs_{n}=ar-ar^{n+1}$. If $r\neq 1$, then this rearranges to give

as required.

(ii) Let $a$, $r\in\mathbb{R}$ with $r\neq 1$. We shall show by induction on $n$ that

Base case: For $n=1$, it is immediate that

which establishes the base case.

Inductive step: Suppose, as an induction hypothesis, that the formula holds for some $n\in\mathbb{N}$. Since $s_{n+1}=s_{n}+ar^{n+1}$, we may apply the induction hypothesis to deduce that

Thus,

This shows that the formula is true for $n+1$, which closes the induction and thus completes the proof.

Observe that

We wish to form the partial fraction decomposition of $\frac{1}{k^{2}+3k+2}$, which means writing

Multiplying through by the denominator,

Since there is no dependence on $k$ on the left-hand side, we must have $A+B=0$ and so $B=-A$. Substituting this into the above, $A=1$ and so $A=1$ and $B=-1$. Thus, we have

From this, we see that $\sum_{k=1}^{\infty}\frac{1}{k^{2}+3k+2}$ is a telescoping series. Using the cancellation, the $n$th partial sum is given by

Since $\frac{1}{n+2}\to 0$ as $n\to\infty$, it follows that $\lim_{n\to\infty}s_{n}=\frac{1}{2}$ and so, by definition, $\sum_{k=1}^{\infty}\frac{1}{k^{2}+3k+2}=\frac{1}{2}$.

By writing

we see that $\sum_{k=1}^{\infty}r^{k}$ is a telescoping series. Using the cancellation,

Since $|r|>1$, the sequence $(r^{n+1})_{n\in\mathbb{N}}$ is unbounded. It follows from the above formula that the sequence of partial sums $(s_{n})_{n\in\mathbb{N}}$ is unbounded and therefore diverges. Thus, by definition, the series diverges.

(i) For all $k\in\mathbb{N}$, by dividing the numerator and dominator by $k$ we have

Since $1/k\to 0$ as $k\to\infty$, it therefore follows from the limit laws that $\frac{k}{k+1}\to 1$ as $k\to\infty$. Consequently, by a result from Worksheet 4, the sequence $((-1)^{k}\frac{k}{k+1})_{k\in\mathbb{N}}$ diverges. Explicitly, the subsequence of odd terms tends to $-1$, whilst the subsequence of even terms tends to $1$. It therefore follows from the subsequence test that $\big{(}(-1)^{k}\cdot\frac{k}{k+1}\big{)}_{k\in\mathbb{N}}$ is a divergent sequence. In particular, it does not converge to 0. Thus, by the $k$th term test, the series $\sum_{k=1}^{\infty}(-1)^{k}\frac{k}{k+1}$ diverges.

(ii) For all $k\in\mathbb{N}$, by dividing the numerator and dominator by $k^{2}$ we have

Since $1/k\to 0$ as $k\to\infty$, it therefore follows from the limit laws that $\frac{k^{2}}{3k^{2}-k}\to\frac{1}{3}$ as $k\to\infty$. Thus, by the $k$th term test, the series $\sum_{k=1}^{\infty}\frac{k^{2}}{3k^{2}-k}$ diverges.

(iii) Since $(3^{k})_{k\in\mathbb{N}}$ is unbounded, $3^{k}\not\to 0$ as $k\to\infty$. Thus, by the $k$th term test, the series $\sum_{k=1}^{\infty}3^{k}$ diverges.

1. We write the $n$th partial sum as

where we denote by $A_{n}$ and $B_{n}$ the $n$th partial sum of $\sum_{k=1}^{\infty}a_{k}$ and $\sum_{k=1}^{\infty}b_{k}$, respectively. By assumption, both $\sum_{k=1}^{\infty}a_{k}$ and $\sum_{k=1}^{\infty}b_{k}$ converge, and so both the sequences $A_{n}$ and $B_{n}$ converge, with $\lim_{n\to\infty}A_{n}=A$ and $\lim_{n\to\infty}B_{n}=B$ for $A=\sum_{k=1}^{\infty}a_{k}$, $B=\sum_{k=1}^{\infty}b_{k}\in\mathbb{R}$. By the limit laws for sequences, the sequence $A_{n}+B_{n}$ converges, with $\lim_{n\to\infty}A_{n}+B_{n}=A+B$. This implies that $\lim_{n\to\infty}s_{n}=A+B$ which, by definition, means that the series $\sum_{k=1}^{\infty}(a_{k}+b_{k})$ converges and $\sum_{k=1}^{\infty}(a_{k}+b_{k})=\sum_{k=1}^{\infty}a_{k}+\sum_{k=1}^{\infty}b_{k}$.

2. We write the $n$th partial of $\sum_{k=1}^{\infty}(\lambda\cdot a_{k})$ sum as

where $A_{n}$ is as defined in the proof of part 1. By assumption, $\sum_{k=1}^{\infty}a_{k}$ converges and so the sequence $A_{n}$ converges, with $\lim_{n\to\infty}A_{n}=A$ for $A=\sum_{k=1}^{\infty}a_{k}\in\mathbb{R}$. By the limit laws for sequence, the sequence $\lambda A_{n}$ converges, with $\lim_{n\to\infty}\lambda A_{n}=\lambda A$. This implies that $\lim_{n\to\infty}s_{n}=\lambda A$ which, by definition, means that the series $\sum_{k=1}^{\infty}(\lambda\cdot a_{k})$ converges and that $\sum_{k=1}^{\infty}(\lambda\cdot a_{k})=\lambda\sum_{k=1}^{\infty}a_{k}$.

Let $(a_{k})_{k\in\mathbb{N}}$ be a sequence such that $\sum_{k=1}^{\infty}a_{k}$ converges. From our earlier observations, given $m\in\mathbb{N}$, the tail $\sum_{k=m+1}^{\infty}a_{k}$ converges and

However, by definition $\lim_{m\to\infty}\sum_{k=1}^{m}a_{k}=\sum_{k=1}^{\infty}a_{k}$ and so, by the limit laws,

as required.

Let $r\in\mathbb{R}$ with $|r|<1$. Let $(t_{n})_{n\in\mathbb{N}}$ denote the partial sums of the series $\sum_{k=1}^{\infty}ar^{k}=\frac{ar}{1-r}$ and $(s_{n})_{n\in\mathbb{N}_{0}}$ the partial sums of $\sum_{k=0}^{\infty}ar^{k}$. It follows that $s_{n}=t_{n}+a$ for all $n\in\mathbb{N}$. Thus, by the limit laws and (3.2), we have

as required.

Let $(a_{k})_{k\in\mathbb{N}}$ and $(b_{k})_{k\in\mathbb{N}}$ satisfy $0\leq a_{k}\leq b_{k}$ for all $k\in\mathbb{N}$ and suppose $\sum_{k=1}^{\infty}a_{k}$ diverges.

Our goal is to show $\sum_{k=1}^{\infty}b_{k}$ diverges; that is, that the sequence of partial sums $\big{(}\sum_{k=1}^{n}b_{k}\big{)}_{n\in\mathbb{N}}$ is unbounded.

Let $M>0$ be given. By the boundedness test from Lemma 3.26, since $\sum_{k=1}^{\infty}a_{k}$ diverges, the sequence of partial sums of $\big{(}\sum_{k=1}^{n}a_{k}\big{)}_{n\in\mathbb{N}}$ is unbounded. In particular, there exists some $n\in\mathbb{N}$ such that

Since $0\leq a_{k}\leq b_{k}$ for all $k\in\mathbb{N}$, we therefore see that

Since $M>0$ was chosen arbitrarily, it follows that the sequence of partial sums $\big{(}\sum_{k=1}^{n}b_{k}\big{)}_{n\in\mathbb{N}}$ is unbounded, as required.

(i) For all $k\in\mathbb{N}$, we have

Since we know $\sum_{k=1}^{\infty}\frac{4}{k^{2}}$ converges by Example 3.29 and the limit laws, it follows by the comparison test that $\sum_{k=1}^{\infty}\frac{3k+1}{4k^{3}-3}$ converges.

(ii) If $k\geq 3$, then $\log_{2}(k+1)\geq\log_{2}(3+1)=\log_{2}4=2$. Hence

Since we know $\sum_{k=3}^{\infty}2^{-k}$ is a convergent geometric series, it follows by the comparison test that the tail $\sum_{k=3}^{\infty}\frac{1}{\log_{2}(k+1)^{k}}$ converges. Hence, $\sum_{k=1}^{\infty}\frac{1}{\log_{2}(k+1)^{k}}$ also converges.

(iii) For $k\geq 4$, we have

Furthermore, if $k\geq 6$, then we have $3\leq k/2$ and so $(k-3)^{4}\geq(k/2)^{4}=k^{4}/16$. From this, we see that

Since we know $\sum_{k=6}^{\infty}\frac{16}{k^{2}}$ converges by Example 3.29 and the limit laws, it follows by the comparison test that the tail $\sum_{k=6}^{\infty}\frac{k^{2}}{k!}$ converges. Thus, $\sum_{k=1}^{\infty}\frac{k^{2}}{k!}$ also converges.

Let $(d_{k})_{k\in\mathbb{N}}$ be a sequence with $d_{k}\in\{0,1,\dots,9\}$ for all $k\in\mathbb{N}$. Then

We know from the geometric series formula that $\sum_{k=1}^{\infty}9\cdot 10^{-k}=1$. Thus, by the comparison test, $\sum_{k=1}^{\infty}d_{k}\cdot 10^{-k}$ converges and

It follows by definition of decimal expansion that $x=0.d_{1}d_{2}d_{3}\ldots$.

(i) Let $(a_{k})_{k\in\mathbb{N}}$ be a positive sequence and suppose the sequence $(r_{k})_{k\in\mathbb{N}}$ given by $r_{k}:=a_{k+1}/a_{k}$ for all $k\in\mathbb{N}$ converges with $r:=\lim_{k\to\infty}r_{k}>1$.

Let $\varepsilon:=\frac{r-1}{2}>0$. By the $\varepsilon$-$N$ definition of a limit, there exists $N\in\mathbb{N}$ such that $|r_{n}-r|<\varepsilon$ for all $n\geq N$. Thus, by the triangle inequality,

For any $k\in\mathbb{N}$ with $k>N$, we may write

Applying (A.5) to each of the factors $r_{n}$ for $N\leq n\leq k-1$, we deduce that

Since $s>1$, we know from Example 3.4 that $\sum_{k=N+1}^{\infty}a_{N}s^{k-N}$ is a divergent geometric series. In light of (A.6), the series $\sum_{k=N+1}^{\infty}a_{k}$ diverges by the comparison test. Finally, since we have shown a tail diverges, we conclude that the whole series $\sum_{k=1}^{\infty}a_{k}$ diverges.

(ii) Let $(a_{k})_{k\in\mathbb{N}}$ be a positive sequence and suppose the sequence $(r_{k})_{k\in\mathbb{N}}$ given by $r_{k}:=a_{k+1}/a_{k}$ for all $k\in\mathbb{N}$ satisfies $\lim_{k\to\infty}r_{k}=\infty$. Then there exists some $N\in\mathbb{N}$ such that $r_{n}>2$ for all $n\geq N$.

Given $k>N$, we may express $a_{k}$ as in (A.6) to deduce that $a_{k}>a_{N}2^{k-N}$. The proof now concludes as in part (i): we know from Example 3.4 that $\sum_{k=N+1}^{\infty}a_{N}2^{k-N}$ diverges and so $\sum_{k=N+1}^{\infty}a_{k}$ diverges by the comparison test. Finally, since we have shown a tail diverges, we conclude that the whole series $\sum_{k=1}^{\infty}a_{k}$ diverges.

(i) Let $a_{k}:=\frac{k^{3}2^{2k}}{3^{k}+1}$ for $k\in\mathbb{N}$. Then

Since $1/k\to 0$ as $k\to\infty$, it follows from the limit laws that $\big{(}1+\frac{1}{k}\big{)}^{3}\to 1$ as $k\to\infty$. Moreover, by multiplying the numerator and denominator by $3^{-k}$, we see that

where we have again used the limit laws. Thus, by one final application of the limit laws,

Since $4/3>1$, by the ratio test, the series $\sum_{k=1}^{\infty}\frac{k^{3}2^{2k}}{3^{k}+1}$ diverges.

(ii) Let $b_{k}:=\frac{5^{k}}{k!}$ for $k\in\mathbb{N}$. Then

Since $0<1$, by the ratio test, the series $\sum_{k=1}^{\infty}\frac{5^{k}}{k!}$ converges.

(iii) Let $c_{k}:=\frac{k!}{5^{k}}$ for $k\in\mathbb{N}$. Then

Thus, by the ratio test, the series $\sum_{k=1}^{\infty}\frac{k!}{5^{k}}$ diverges.

(iv) Let $d_{k}:=\frac{k^{10}}{(2k-1)!}$ for $k\in\mathbb{N}$. Then

Since $1/k\to 0$ as $k\to\infty$, it follows from the limit laws that $\big{(}1+\frac{1}{k}\big{)}^{10}\to 1$ as $k\to\infty$. On the other hand, $0\leq\frac{1}{2k(2k+1)}\leq\frac{1}{k}$ for all $k\in\mathbb{N}$ and so, by the squeeze theorem, $\frac{1}{2k(2k+1)}\to 0$ as $k\to\infty$. Thus, by another application of the limit laws,

Since $0<1$, by the ratio test, the series $\sum_{k=1}^{\infty}\frac{k^{10}}{(2k-1)!}$ converges.

Let

so that $(s_{n})_{n\in\mathbb{N}}$ and $(t_{n})_{n\in\mathbb{N}}$ are the sequences of partial sums of the series $\sum_{k=1}^{\infty}a_{k}$ and $\sum_{k=0}^{\infty}2^{k}a_{2^{k}}$, respectively. Consider the subsequence $(s_{2^{n}})_{n\in\mathbb{N}}$ and partition the terms of $s_{2^{n}}$ into ‘blocks’

where we have used the hypothesis that $(a_{n})_{n\in\mathbb{N}}$ is nonincreasing. Therefore

If the condensed series $\sum_{k=0}^{\infty}2^{k}a_{2^{k}}$ converges, then the sequence of partial sums $(t_{n})_{n\in\mathbb{N}}$ is bounded. By (A.8), the subsequence of partial sums $(s_{2^{n}})_{n\in\mathbb{N}}$ is therefore also bounded, so there exists some $M>0$ such that $0\leq s_{2^{n}}\leq M$ for all $n\in\mathbb{N}$. But then $0\leq s_{n}\leq M$ for all $n\in\mathbb{N}$ and so the complete sequence of partial sums $(s_{n})_{n\in\mathbb{N}}$ is also bounded. Indeed, given $n\in\mathbb{N}$, we have $n\leq 2^{n}$ and since the terms of the sequence $(a_{k})_{k\in\mathbb{N}}$ are non-negative, it follows that $0\leq s_{n}\leq s_{2^{n}}\leq M$. Consequently, $\sum_{k=1}^{\infty}a_{k}$ converges by the boundedness test from Lemma 3.26.

(i) Since the sequence of terms $\big{(}\frac{1}{k(\log_{2}k)^{2}}\big{)}_{k=2}^{\infty}$ is nonincreasing and non-negative, we can apply the condensation test. The condensed series is given by

which know converges. Thus, by the condensation test, $\sum_{k=2}^{\infty}\frac{1}{k(\log_{2}k)^{2}}$ converges.

(ii) Since the sequence of terms $\big{(}\frac{1}{(\log_{2}k)^{\log_{2}k}}\big{)}_{k=2}^{\infty}$ is nonincreasing and non-negative, we can apply the condensation test. The condensed series is given by

We claim this series converges. Once we have proved the claim, it follows by the condensation test that $\sum_{k=2}^{\infty}\frac{1}{(\log_{2}k)^{\log_{2}k}}$ converges.

We can show $\sum_{k=1}^{\infty}\frac{2^{k}}{k^{k}}$ converges using the ratio test. Indeed, let $a_{k}:=\frac{2^{k}}{k^{k}}$ for all $k\in\mathbb{N}$ so that

Observe that

Thus, by the squeeze theorem, $a_{k+1}/a_{k}\to 0$ as $k\to\infty$. Since $0<1$, by the ratio test, the series $\sum_{k=1}^{\infty}\frac{2^{k}}{k^{k}}$ converges, as claimed.

(i) Observe that

Since $5/3>1$, it follows by the $p$-series test that $\sum_{k=1}^{\infty}\frac{1}{k^{5/3}}$ converges. Hence, by the comparison test, $\sum_{k=1}^{\infty}\frac{k^{1/3}}{k^{2}+10}$ converges.

(ii) For $k\in\mathbb{N}$, observe that

Thus,

Since $1+1/100>1$, it follows that $\sum_{k=1}^{\infty}\frac{1}{k^{1+1/100}}$ converges by the $p$-series test. Hence, $\sum_{k=1}^{\infty}\frac{\sqrt{k^{2}+1}-k}{k^{1/100}}$ converges by the comparison test.

(i) Recall that our goal is to show

Let $m$, $n\in\mathbb{N}$ with $n>m$ and suppose $m$ and $n$ have the same parity: that is, either both are odd or both are even. Then

Since, by hypothesis, $(a_{k})_{k\in\mathbb{N}}$ is nonincreasing, $a_{k-1}-a_{k}\geq 0$ for all $k\geq 2$. In particular, the above shows that $(-1)^{m+1}\sum_{k=m+1}^{n}(-1)^{k}a_{k}$ is a sum of non-negative terms and therefore

Using (A.10), we now group the terms as

Arguing as before, $a_{k+1}-a_{k}\leq 0$ for all $k\in\mathbb{N}$. On the other hand, since $(a_{k})_{k\in\mathbb{N}}$ is non-negative, $-a_{n}\leq 0$. Combining these observations, we see that

which verifies (A.9) in this case.

(ii) Now suppose $m$ and $n$ are have different parity. Then

is again a sum of non-negative terms. Thus, (A.10) continues to hold in this case. Using (A.10), we now group the terms as

where we have again used $a_{k+1}-a_{k}\leq 0$ for all $k\in\mathbb{N}$. This completes the proof.

(i) Since the sequence $(1/k^{3})_{k\in\mathbb{N}}$ is nonincreasing, non-negative and $1/k^{3}\to 0$ as $k\to\infty$, the alternating sign test implies that $\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^{3}}$ converges.

(ii) Since the sequence $(1/\log_{2}(k+1))_{k\in\mathbb{N}}$ is nonincreasing, non-negative and $1/\log_{2}(k+1)\to 0$ as $k\to\infty$, the alternating sign test implies that $\sum_{k=1}^{\infty}\frac{(-1)^{k}}{\log_{2}(k+1)}$ converges.

(iii) Recall that

for all $k\in\mathbb{N}$, so that $(\sqrt{k+1}-\sqrt{k})_{k\in\mathbb{N}}$ is nonincreasing, non-negative and $\sqrt{k+1}-\sqrt{k}\to 0$ as $k\to\infty$. Thus, the alternating sign test implies that $\sum_{k=1}^{\infty}(-1)^{k}(\sqrt{k+1}-\sqrt{k})$ converges.

(iv) An easy application of the limit laws shows that $\frac{2k}{3k+1}\to\frac{2}{3}$ as $k\to\infty$. It then follows from a result from Worksheet 4 that $\big{(}(-1)^{k}\cdot\frac{2k}{3k+1}\big{)}_{k\in\mathbb{N}}$. Explicitly, the subsequence of odd terms tends to $-2/3$, whilst the subsequence of even terms tends to $2/3$; it therefore follows from the subsequence test that $\big{(}(-1)^{k}\cdot\frac{2k}{3k+1}\big{)}_{k\in\mathbb{N}}$ diverges. Hence, by the $k$th term test for convergent series, the series $\sum_{k=1}^{\infty}(-1)^{k}\cdot\frac{k}{k+1}$ diverges.

(i) Taking absolute values,