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$$\sum _{k=1} ^\infty \frac{(-1)^k}{k}$$

I know this question has been answered a few times but my professor has not taught alternating series test yet or anything other than ratio test, root test and comparison test where $a_i \geq 0$ for every $i \in \mathbb N$ and $\sum_{i=1} ^\infty a_i$ converges and if $|b_i| \leq a_i$ for every i then $\sum_{i=1} ^\infty b_i$ converges absolutely.

So here's my attempt using Cauchy criterion.

What we know: We say that the series $\sum _{i=1}^\infty a_i$ converges if the sequence of partial sums $(S_i)_i$$_\in$$_\mathbb N$ converges.

From Cauchy criterion, $(S_i)_i$$_\in$$_\mathbb N$ converges if and only if it is a Cauchy sequence.

It is quite obvious that $$\lim_{k\to\infty} S_k = \sum _{k=1}^\infty \frac{(-1)^k}{k}$$.

I denote $\lim_{k\to\infty} S_k = S$

Suppose ($S_k$) is convergent. Then $\forall \epsilon \gt 0, \exists N \in \mathbb N$ such that $\forall n \geq N,$

$|S_n - S|$ = $\vert \sum_{k=n+1} ^\infty \vert$ $\lt \epsilon$

I am stuck here. Is it possible to find such N for all $\epsilon \gt 0$ to hold

$\vert \sum_{k=n+1} ^\infty \vert$ $\lt \epsilon$

to be true?

(If yes, then the sequence ($S_k$) is convergent so $\sum_{k=1} ^\infty \frac{(-1)^k}{k}$ is convergent by the definition but I don't quite understand if we can always find such N)

edit: I don't think ratio test or root test are applicable to solve this and is the alternating series test the only way to solve this problem?

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Hint. Let $$ s_n=\sum_{k=1}^n\frac{(-1)^k}{k} $$ Then observe that $$ s_1<s_3<\cdots<s_{2n-1}<s_{2n+1}<s_{2n+2}<s_{2n}<s_{2n-2}<\cdots<s_4<s_2 $$ Hence $a_n=s_{2n-1}$ is increasing and upper bounded, by $s_2$, while $b_n=s_{2n}$ is decreasing and lower bounded by $s_1$. Hence both converge, and since $a_n<b_n$, then $$ \lim a_n\le \lim b_n $$ But $b_n-a_n=\frac{1}{2n}\to 0$, and hence $$ \lim a_n= \lim b_n $$

Note. Inevitably, the idea of the proof of Alternating series test is used in the above proof.

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    $\begingroup$ Of course, this is just the proof of the alternating series test. :) $\endgroup$ – Ted Shifrin Nov 4 '18 at 17:08
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    $\begingroup$ Isn’t it the same argument to prove the alternating series test? $\endgroup$ – gimusi Nov 4 '18 at 17:27
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    $\begingroup$ Then why don’t use directly the test! Uour answer is fine but you should explain that I think to let know the asker that this way is equivalent to use the alternating series test. $\endgroup$ – gimusi Nov 4 '18 at 17:30
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    $\begingroup$ Yes but here we are using that! The idea is the same, it is exactly the test. Maybe the question should be show convergence using the proof for alternating series test. $\endgroup$ – gimusi Nov 4 '18 at 17:35
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    $\begingroup$ I agree with Gimusi, this is the alternating test, even though you prove it from scratch. $\endgroup$ – Yves Daoust Nov 4 '18 at 21:56
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Not only can we show that the series of interest converges, we can also evaluate it using elementary tools only. We now present two approaches.


METHODOLOGY $1$:

Noting that $\int_0^1 x^{n-1}\,dx=\frac1n$, we can write

$$\begin{align} \sum_{n=1}^N \frac{(-1)^n}{n}&=\sum_{n=1}^N (-1)^n\int_0^1 x^{n-1}\,dx\\\\ &=-\int_0^1 \sum_{n=1}^N (-x)^{n-1}\,dx\\\\ &=-\int_0^1 \frac{1-(-x)^N}{1+x}\,dx\\\\ &=-\log(2)+(-1)^N\int_0^1 \frac{x^N}{1+x}\,dx \end{align}$$

Since $\left|\frac{x^N}{1+x}\right|\le x^N$, we have the estimate

$$\left|\int_0^1 \frac{x^N}{1+x}\,dx\right|\le \frac{1}{N+1}$$

Therefore,

$$\lim_{N\to\infty }\sum_{n=1}^N \frac{(-1)^n}{n}=-\log(2)$$


METHODOLOGY $2$:

$$\begin{align} \sum_{n=1}^{2N}\frac{(-1)^n}{n}&=\sum_{n=1}^N \frac1{2n}-\sum_{n=1}^N \frac1{2n-1}\\\\ &=\sum_{n=1}^N \frac1{2n}-\left(\sum_{n=1}^{2N}\frac1n-\sum_{n=1}^N\frac1{2n}\right)\\\\ &=-\sum_{n=N+1}^{2N}\frac1n\\\\ &=-\sum_{n=1}^N \frac{1}{n+N}\\\\ &=-\frac1N \sum_{n=1}^N\frac1{1+(n/N)} \end{align}$$

The last expression is the Riemann Sum for $-\int_0^1 \frac1{1+x}\,dx=-\log(2)$ as expected.


If these methodologies are not quite the way forward that the OP is seeking, then we simply note

$$\begin{align} \left|\sum_{n=1}^{2N}\frac{(-1)^n}{n}\right|&=\left|\sum_{n=1}^N \frac1{2n}-\sum_{n=1}^N \frac1{2n-1}\right|\\\\ &=\sum_{n=1}^N \frac{1}{2n(2n-1)}\\\\ &\le \frac12 \sum_{n=1}^N \frac1{n^2} \end{align}$$

and the series of interest converges by comparison with the series $\sum_{n=1}^\infty \frac{1}{n^2}$ (Note that $\sum_{n=2}^\infty \frac{1}{n^2}\le \sum_{n=2}^\infty \frac{1}{n(n-1)}=1$).

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  • $\begingroup$ @TUC Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Nov 5 '18 at 15:24
  • $\begingroup$ Three answers to the actual question! +1 $\endgroup$ – RRL Nov 7 '18 at 7:26
  • $\begingroup$ @rrl Thank you for the up vote! Much appreciated. $\endgroup$ – Mark Viola Nov 7 '18 at 13:22
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We can process the terms in pairs because two successive partial sums differ in $\pm\dfrac1n$, which tends to zero.

Now let us study the series with general term

$$\frac1{2n}-\frac1{2n+1}=\frac1{(2n+1)2n}\le\frac1{4n^2}.$$

For this upper bound, we will group the second and third terms, then the fourth to seventh, and so on, each time doubling the length.

$$4S=1+\left(\frac1{2^2}+\frac1{3^2}\right)+\left(\frac1{4^2}+\frac1{5^2}+\frac1{6^2}+\frac1{7^2}\right)+\cdots \\\le1+\left(\frac1{2^2}+\frac1{2^2}\right)+\left(\frac1{4^2}+\frac1{4^2}+\frac1{4^2}+\frac1{4^2}\right)+\cdots \\=1+\frac12+\frac14+\cdots,$$ a well known convergent series.

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  • $\begingroup$ Mark Viola gives a much simpler argument: you can bound with a telescoping series. The above method works for all exponents $>1$ and can be adapted to prove divergence of the Harmonic series. $\endgroup$ – Yves Daoust Nov 4 '18 at 22:17

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