How to prove: if a part of the series does not go to 0 then it diverges

Question

Sorry for the title, isn't precise.

I mean to say, if we have a sequence $(a_n)$, that defines $x_n = \sum_{k= n+1}^{2n}a_k$. So, what I want the prove is, if $ lim \ x_n =L$, such that $L \ne 0$ the serie $\sum_{n=0}^{\infty}a_n$ diverges.

I try to apply the Cauchy criterion for series, that says: for a series to converge it is necessary and sufficient that $\forall \epsilon \gt0 \ \exists n_0$ such that $|a_n + a_{n+1}+...+a_{n+p}|\lt\epsilon$.

Then I wrote $x_n = s_{2n} - s_n$, such that $s_n$ is partial sums of $\sum a_n$ so I have that $|s_{2n} - s_n - L|\lt \epsilon$ (definition of limit). And I don't where to go, please help.

Thank you.

Answer 1

Assume that $\displaystyle\sum_{n=0}^{\infty}a_{n}$ exists, then by denoting $s_{n}=\displaystyle\sum_{k=0}^{n}a_{k}$, then $\{s_{n}\}$ is Cauchy, which means that $s_{n}-s_{m}\rightarrow 0$ as $n,m\rightarrow\infty$, in particular, we have $s_{2n}-s_{n}\rightarrow 0$ as $n\rightarrow\infty$.

Answer 2

Hint: Look at the sequence $$ s_1, s_2, s_4, s_8, \ldots, s_{2^n}, \ldots $$ Observe that these are the partial sums of a certain subsequence of $x_i$s.

Answer 3

Remember if $\{x_n\}\to L$ converges then any subsequence $\{x_{n_i}\}$ converge to $l$ also.

Let $y_n = x_{2^n}$. Then $y_n \to L$ converges.

And $\sum_{k=0}^\infty y_n = \sum_{k=0}^\infty \sum_{j= 2^k + 1}^{2^{k+1}} a_j = \sum_{n=0}^\infty a_n$.

And as $y_n\to L \ne 0$ then $\sum_{k=0}^\infty y_n$ diverges.