Prove that $\lim_{n \to \infty} \sum_{k=1}^n \frac{(-1)^k}{k}$ exists and is finite.
Attempt: Suppose $\{x_n\}$ is real sequence, and $x_n = \frac{(-1)^k}{k}$.
I know if I prove that it is Cauchy, then it converges.
Then, by definition of Cauchy: for every $\epsilon >0$ there is a natural number $N$ such that $m,n \geq N$ implies $|x_m – x_n| < \epsilon$.
Thus, using $m \geq n \geq N$ implies $|\sum_{k=m}^n (x_m-x_{n-1})| < \epsilon$.
Then if I show $|\sum_{k=m}^n (x_m-x_{n-1}) | $ converges, then I can conclude so does $\{x_n\}$ thus the limit exists and is finite.
However, I am having trouble. Please can anyone please help me? Anything feedback/hint would be really appreciated.
Thank you.
Best Answer
It appears that you're trying to show the convergence of the sum $\sum_{i=1}^{\infty} x_i$ by considering whether the series $x_i$ converges. This doesn't work- consider that $1, \frac{1}2, \frac{1}3, \frac{1}4,\ldots$ converges, the sum of those terms is the harmonic series and does not converge. However, the argument you're trying to apply to this problem would also seem to tell us that the harmonic series does converge - so the argument must be wrong. I'm also quite confused how you get to the equation $|\sum_{k=m}^n (x_m-x_{n-1})|<\varepsilon$ - I don't think that follows.
What you really want to do is to define a new series, let's call it $s$ defined as $$s_n=\sum_{i=1}^n x_i$$ or, equivalently, $$s_n=s_{n-1}+x_n.$$ You need to show that this is Cauchy. You can do this for the particular case where $x_i$ is $\frac{(-1)^i}{i}$, but it suffices to see that:
Then the critical thing to note is that, if you choose any $k$, then it holds that $$s_{2k+1}<s_{2k+3}<s_{2k+4}<s_{2k+2}.$$ This is easy enough to check by noting things like the fact that $s_{2k+3}-s_{2k+1}$ is equal to $x_{2k+2}+x_{2k+3}$ - which is a positive number plus a smaller negative, and hence positive. But, the critical thing of note here is that it easily follows from the above that for any $M>2k$, all the $s_M$ are in the interval $[s_{2k+1},s_{2k+2}]$ and that this interval has length $x_{2k+2}$ - meaning no two element within are more than $x_{2k+2}$ apart.
The proof that $s_k$ is Cauchy should be easy enough from there.