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I need to prove this sum is diverges/convergent/conditional convergent , but I am pretty sure it is converging to a value:

$$\sum_{k=1}^{\infty} \frac{(1+\frac{1}{k})^{k^a}}{k!}$$
For some constant:

$a >0 , a \in \mathbb{R}$

I tried to prove it using:

$\frac{(1+\frac{1}{n})^{k^a}}{k!} \leq \frac 1k \rightarrow \frac{k \cdot (1+\frac{1}{n})^{k^a}}{(k)!} \leq 1 \rightarrow \frac{(\frac{k+1}{k})^{k^a}}{(k-1)!} \rightarrow$ now I divide both by $k^{k^a}$ and get:
$\frac{(k+1)^{k^a}}{k^{k^a}} \cdot \frac{1}{(k-1)!}$ and I check the $\text{limit}$ as the first term in this product $k \rightarrow \infty$ and get $0$

And thus it is divergent because $\frac{(1+\frac 1k)^{k^a}}{k!} \leq \frac 1n$ (Harmonic divergent)

How can I solve this for any given $a \in \mathbb{R} , ~~ a > 0 $ ?

Thank you very much!

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1 Answer 1

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For $a \leq 1$, the numerator $$ (1+\frac{1}{k})^{k^a} < e $$ because $a_n = (1+\frac{1}{n})^n$ is an increasing series that converges to $e$. So the series is convergent.

EIDT: For $a=2$, the upper bound in $$ (1+\frac{1}{k})^{k^a}<(1+\frac{1}{k})^{k^{1+1}}<e^k $$ The series converges. because the sum you have converges for all $b_k = |x|^k$/k!

For $a>2,$ consider ratio test with $$ a_k = \frac{(1+\frac{1}{k})^{k^a}}{k!} $$ Then, take $|\frac{a_{k+1}}{a_k}|$. Write the expressions in the numerator and denominator as $a=e^{\log a}$, then use Taylor series expansion to get $$ \frac{e^{n^{a-1}}(1+\frac{1}{n})^{a-1} - 1}{n+1}>\frac{e^{n^{a-2}}(a-1)}{n+1} $$ which diverges for $a>2$

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  • $\begingroup$ So it is conditionally converging when $a \leq 1$ ? what about when $a geq 1$ ? Thank you! $\endgroup$
    – CSch of x
    Jul 11, 2020 at 21:09
  • $\begingroup$ see the edit pls $\endgroup$
    – Alex
    Jul 11, 2020 at 21:31
  • $\begingroup$ Thanks so much but I did not really understand the set of $a$ that this sum converges for , and I am sorry for being annoying, so this sum converges for all $a \in \mathbb{R} ~~ a > 0$ ? or only when $a \leq 1$ and $a>2$ it converges? and thus for $a \in (1,2)$ it diverges $\endgroup$
    – CSch of x
    Jul 11, 2020 at 21:38
  • $\begingroup$ It converges for $a \leq 1$, but for $a>1$ I'm not sure. The ratio test I did was wrong $\endgroup$
    – Alex
    Jul 11, 2020 at 21:40
  • $\begingroup$ But why though? because as you said is $\frac{e^{n^{a-2}}(a-1)}{n+1}$ and if $a >1 \wedge a <2$ then we can't tell something about it $\endgroup$
    – CSch of x
    Jul 11, 2020 at 22:03

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