# Prove that $\sum a_k$ convergent implies $ka_k \to 0.$

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.

Max

• Cauchy's convergence test, or Cauchy's condensation test? – zhw. Dec 14 '17 at 21:43
• @zhw. cauchy's convergence test – MLK Dec 14 '17 at 22:21
• Do the instructions require you to use the cauchy convergence test explicitly? – Zackkenyon Dec 14 '17 at 23:24
• 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 – MLK Dec 14 '17 at 23:25

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 $k>M$.