Question: Show that the series $\sum_{n=0}^\infty \frac{(-1)^n}{n}$ converges as follows. Let Sn denote the nth partial sums. Show that S2n−1 is a bounded monotone sequence and therefore convergent. Then show that Sn actually converges by noting that the difference between S2n and S2n−1 is small.
Let an = $\frac{(-1)^n}{n}$, Sn = $\sum_{n=0}^\infty$ an
a1 = -1 --------------------- a2 = 1/2
a3 = -1/3 ------------------ a4 = 1/4
a5 = -1/5 ------------------ a6 = 1/6
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S2n-1 = $\sum_{n=0}^\infty$ a2n-1
S1 = -1
S2 = -1/3
S3 = -1/5
Claim: S2n-1 is monotone increasing
S1 < S2
S2 < S3
Prove using induction:
Let P(n): S2n-1 < S2n
P(1): S1 < S2 : -1 < -1/3 (True)
P(2): S3 < S4 : -1/5 < -1/7 (True)
This is what I have got so far, can anyone help me out because I'm lost, the induction is confusing for me, thank you very much.
The series converges according to the Leibniz's alternating series test. https://en.wikipedia.org/wiki/Alternating_series_test .
The OP's method also works as follows: \begin{align} b_n&=S_{2n}=\sum_{k=1}^{n}\left(-\frac1{2k-1}+\frac1{2k}\right)=b_{n-1}-\frac1{(2n-1)2n}<b_{n-1}\\ b_n&=\sum_{k=1}^{n}\frac{-1}{(2k-1)2k}>\sum_{k=1}^{n}\frac{-1}{(2k-1)^2}>-1-\sum_{k=2}^{n}\frac{1}{(2k-2)^2}=-1-\frac14\sum_{k=1}^{n-1}\frac{1}{k^2}\\ &>-1-\frac{\pi^2}{24}\\\\ \text{So }b_n&\text{ is decreasing but lower bounded. Similarly,}\\\\ c_n&=S_{2n+1}=-1+\sum_{k=1}^{n}\left(\frac1{2k}-\frac1{2k+1}\right)=c_{n-1}+\frac1{2n(2n+1)}>b_{n-1}\\ c_n&=-1+\sum_{k=1}^{n}\frac{1}{2k(2k+1)}<-1+\sum_{k=1}^{n}\frac{1}{(2k)^2}=-1+\frac14\sum_{k=1}^{n}\frac{1}{k^2}<-1+\frac{\pi^2}{24}\\\\ \text{So }c_n&\text{ is increasing but upper bounded.}\\\\ \end{align} Therefore, both $S_{2n}$ and $S_{2n+1}$ converges. And as both of those converge, \begin{align} 0&=\lim_{n\to\infty}\left(S_{2n}-S_{2n+1}\right)=\lim_{n\to\infty}S_{2n}-\lim_{n\to\infty}S_{2n+1}\\ \therefore&\lim_{n\to\infty}S_{2n}=\lim_{n\to\infty}S_{2n+1} \end{align} So $S_n$ converges.