Let $ S_1=\displaystyle\sum_{n=1}^{\infty}a_n$ be a positive convergant series, prove that $ S_2=\displaystyle\sum_{n=1}^{\infty}2(a_n)^3$ converges as well.

We have $\exists l :\forall \epsilon>0 : \exists N\in \mathbb N : \forall n > N : |S_1-l|<\epsilon$

We want to show that $\forall \epsilon>0 : \exists N'\in \mathbb N : \forall n > N' : |S_2-L|<\epsilon$

Suppose that $S_2$ does not converge, then cauchy criterion will be false about it:

$\forall n> N' :p\ge 1 : |2a^3_{n+1}+2a^3_{n+2}+...+2a^3_{n+p}|<\epsilon$

$a_n$ is positive, dividing by a constant and applying cube root won't change its convergence so the following statement would have to be false:

$\forall n> N :p\ge 1 : |a_{n+1}+a_{n+2}+...+a_{n+p}|<\epsilon$

So $S_1$ diverges, contradiction.

Another way:

Using limit comparision test:

$\displaystyle\lim_{n\to\infty} \frac {a_n} {b_n}=l \ $ if $\ l=0$ then $\sum a_n =l \Rightarrow \sum b_n = l$

So: $\displaystyle\lim_{n\to\infty} \frac {a_n} {2a_n^3}=\frac {1} {2a_n^2}=0$

So: $S_1=l \Rightarrow S_2=l$

Are these approaches alright ?

  • $\begingroup$ Limit comparision test requires $a_n>0$ and "applying cube root won't change its convergence" not even make sense. $\endgroup$ Apr 10, 2014 at 9:55
  • $\begingroup$ @Martín-BlasPérezPinilla it is, forgot to add this info. About the root, is it true to say that applying root on a converging series won't affect its convergence ? $\endgroup$
    – GinKin
    Apr 10, 2014 at 10:00
  • 1
    $\begingroup$ $\sum \frac{1}{n^2}$ converges but $\sum \sqrt{\frac{1}{n^2}} = \sum \frac{1}{n}$ diverges... $\endgroup$ Apr 10, 2014 at 10:02

1 Answer 1


Much easier: $\sum_{n=1}^{\infty}a_n$ convergent $\implies\lim_{n\to\infty}a_n=0\implies$ $a_n\le 1$ for $n$ large enough $\implies 2a_n^3\le 2a_n$ for $n$ large enough $\implies$ $\sum_{n=1}^{\infty}2a_n^3$ convergent.


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