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I have to prove the following statement:

$(a_k)_{k \in \mathbb{N}}$ is a monotonously falling null sequence. Prove (with the help of Cauchy's convergence test):

If $\sum_{k=0}^\infty a_k \thinspace convergent \implies \lim\limits_{k \rightarrow \infty}{ka_k}= 0 $

My guess was to rewrite the both statements above into their specific form. What I mean with that is, that I rewrote $\sum_{k=0}^\infty$ in cauchy's convergence test form $$|\sum_{k=m}^n a_k|=|a_m+a_{m+1}+a_{m+2}+...+a_n|$$ where $n \geq m \geq N \in \mathbb{N}$

and to rewrite the limes in the $\epsilon-N$ Definition: $$\lim\limits_{k \rightarrow \infty}{ka_k}=|ka_k - 0| < \epsilon $$ where $\epsilon > 0$ and $k \geq N \in \mathbb{N}$.

Sadly, I don't see how I go from there or if this is the right approach.

Does anyone have an idea what would be a good way to go from there? I appreciate every suggestion.

Note: I don't have a definition how $a_n$ looks like. It has to be a complete general prove.

Thank you in advance,

Max

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  • $\begingroup$ Cauchy's convergence test, or Cauchy's condensation test? $\endgroup$ – zhw. Dec 14 '17 at 21:43
  • $\begingroup$ @zhw. cauchy's convergence test $\endgroup$ – MLK Dec 14 '17 at 22:21
  • $\begingroup$ Do the instructions require you to use the cauchy convergence test explicitly? $\endgroup$ – Zackkenyon Dec 14 '17 at 23:24
  • $\begingroup$ I rather think it is meant as a very very clear hint. Probably because the Cauchy Convergence Test is supposed to be the best approach $\endgroup$ – MLK Dec 14 '17 at 23:25
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By the Cauchy convergence test, for every $\varepsilon>0$, there exists $N$ such that, for every $n>N$ and every $p$, $$ |a_n+a_{n+1}+\dots+a_{n+p}|<\varepsilon/2 $$ and we can disregard the absolute value because all terms are positive.

Take $M=2N+1$ and $k>M$. If $k$ is even, then you have $k/2>N$, so $$ a_{k/2}+\dots+a_k<\varepsilon/2 $$ Since the sequence is decreasing, we get, as the number of terms is $1+k/2$, $$ \frac{k}{2}a_k<\frac{\varepsilon}{2} $$ If $k$ is odd, then $(k-1)/2>N$, so $$ a_{(k-1)/2}+\dots+a_k<\varepsilon/2 $$ and therefore, as the number of terms is $1+(k-1)/2$, $$ \left(1+\frac{k-1}{2}\right)a_k<\frac{\varepsilon}{2} $$ which implies $$ \frac{k}{2}a_k<\left(1+\frac{k-1}{2}\right)a_k<\frac{\varepsilon}{2} $$ In both cases, $|ka_k|<\varepsilon$. Thus this holds for every $k>M$.

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  • $\begingroup$ Awesome. Thank you. This looks absolutely easy and smooth. I appreciate the help $\endgroup$ – MLK Dec 14 '17 at 23:08

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