In the problems below it's asked for which $r \in \Bbb R$ the series converges.

$$ a)\quad\sum_{k=0}^\infty \left( \left(\sum_{l=1}^k \frac1l\right) r^k\right) $$

$$ b)\quad\sum_{k=0}^\infty \left( 2^{k^2} r^k \right) $$

In the first problem I use the summation formula of a finite geometric series $\sum_{k=0}^n a r^k = \frac{a-ar^{n+1}}{1-r}$ and rewrite it as $\lim_{n\to 0} \sum_{k=0}^n ar^k = \lim_{n\to 0} \frac{a-ar^{n+1}}{1-r}$. Now by defining $a:=\sum_{l=1}^k \frac1l$ and using $\lim_{n\to 0} \frac{a-ar^{n+1}}{1-r}$ I can use the case analysis to find for which $r \in \Bbb R$ the series converges.

Case 1: If $|r| > 1$, then the term $r^{n+1}$ grows rapidly and the whole term goes towards infinity and diverges.

Case 2: If $|r| = 1$, then the term becomes undefined.

Case 3: If $ 0 < |r| < 1$, then the term $r^{n+1}$ gets smaller and smaller and loses its affect on the whole term and the series converges towards $\frac{a}{1-r}$.

By problem b because of $k^2$ and the rapid growth of $2^{k^2}$ towards infinity any value of $r$ either pushes the series even more towards infinity (for $|r| > 1$) or at most slows down the divergence (for $|r| < 1$).

But my question is if this approach is correct and if yes, is it acceptable as a proof or do I have to use criteria like Root test or Leibniz test or etc. to show the convergence?

Thank you for your help.

  • $\begingroup$ You say $\sum_{k=0}^n a r^k = \frac{a-ar^{n+1}}{1-r}$. But that doesn't generally work if the thing you're calling "$a$" depends on $k$. ${}\qquad{}$ $\endgroup$ May 11 '14 at 16:49
  • $\begingroup$ You mean that $\sum_{l=1}^k \frac1l$ cannot be defined as $a$ because it is not a constant? $\endgroup$
    – Infinity
    May 11 '14 at 16:55
  • $\begingroup$ One could write $a=\text{that number}$, but then one could not validly apply that identity. Thus it would be imprudent to use a notation that doesn't make it clear that the thing you're defining depends on $k$. ${}\qquad{}$ $\endgroup$ May 11 '14 at 16:58

The first series yields to a ratio test: $$ \lim_{k\to\infty}\frac{\sum_{\ell=1}^{k+1} \frac1\ell |r^{k+1}|}{\sum_{\ell=1}^k \frac1\ell |r^k|} = |r|\lim_{k\to\infty} \left( 1 + \frac{1/(k+1)}{\sum_{\ell=1}^k 1/\ell} \right) $$

In the fraction $\dfrac{1/(k+1)}{\sum_{\ell=1}^k 1/\ell}$, the numerator approaches $0$ while the denominator grows. Therefore, that fraction approaches $0$.

Hence the series converges if $-1<r<1$ and diverges if $|r|>1$.

What happens when $r=1$ or $r=-1$ could bear closer examination.

Now let's look at the second series: $$ \lim_{k\to\infty}\frac{2^{(k+1)^2} |r|^{k+1}}{2^{k^2} |r|^k} = \lim_{k\to\infty} 2^{2k+1}|r| =\infty\text{ unless }r=0. $$ So the series diverges unless $r=0$.


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