# Converging / Diverging sum with a constant power:

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!

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 $$(1+\frac{1}{k})^{k^a}<(1+\frac{1}{k})^{k^{1+1}} The series converges. because the sum you have converges for all $$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 $$a>2$$

• So it is conditionally converging when $a \leq 1$ ? what about when $a geq 1$ ? Thank you! Jul 11, 2020 at 21:09
• see the edit pls
– Alex
Jul 11, 2020 at 21:31
• 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 Jul 11, 2020 at 21:38
• It converges for $a \leq 1$, but for $a>1$ I'm not sure. The ratio test I did was wrong
– Alex
Jul 11, 2020 at 21:40
• 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 Jul 11, 2020 at 22:03