# Thread: Convergence value of sum_n=0:Inf ( x^n / n! )^k

1. ## Convergence value of sum_n=0:Inf ( x^n / n! )^k

Is there any way to compute analytically the convergence value of the series:

$\sum_{n=0}^{+\infty}\left(\frac{\lambda^n}{n!}\right)^k$

We know that for k=1, the convergence value is $e^\lambda$.
For k->+Inf the series converges to 2.
For k>1 it is possible to verify that the series converges. The problem is to determine analytically this value.
My questions are:
1. Is there any way to do this?
2. Is there any relation with the Napier number?

2. ## Re: Convergence value of sum_n=0:Inf ( x^n / n! )^k

For k->+Inf the series converges to 2.
For lambda=3, the sum is $(\frac{1}{1})^k + (\frac{3}{2})^k + (\frac{9}{6})^k + (\frac{27}{24})^k + \sum_i a_i^k$ where all ai<1. With k -> inf, this sum vanishes, but the first terms and their sum diverge.

In the limit k->inf:
With lambda=-2, it diverges (oscillates between 0 for odd k and 2 for even k)
With -2<lambda<2, it converges to 1.
With lambda=2, it converges to 2.
With lambda>2, it diverges

With lambda< -2, I would expect divergence, too.

The series itself converges for every k>1, I agree. For every lambda, only a finite number of terms is larger than 1 and with k=1 it converges, therefore taking the k'th power does not change convergence. However, the limit k->inf depends on lambda.

The problem is to determine analytically this value.
No useful idea, sorry.

3. ## Re: Convergence value of sum_n=0:Inf ( x^n / n! )^k

Originally Posted by mfb
For lambda=3, the sum is $(\frac{1}{1})^k + (\frac{3}{2})^k + (\frac{9}{6})^k + (\frac{27}{24})^k + \sum_i a_i^k$ where all ai<1. With k -> inf, this sum vanishes, but the first terms and their sum diverge.

In the limit k->inf:
With lambda=-2, it diverges (oscillates between 0 for odd k and 2 for even k)
With -2<lambda<2, it converges to 1.
With lambda=2, it converges to 2.
With lambda>2, it diverges

With lambda< -2, I would expect divergence, too.

The series itself converges for every k>1, I agree. For every lambda, only a finite number of terms is larger than 1 and with k=1 it converges, therefore taking the k'th power does not change convergence. However, the limit k->inf depends on lambda.

No useful idea, sorry.
Thnx for your reply. There was a lapsus in my first message. When I said that the series converges to 2 for k->Inf I ment for lambda = 1, so we have the following limit:

$\lim_{k\to+\infty}\sum_{n=0}^{+\infty}\left(\frac{ 1}{n!}\right)^k$

by putting in evidence the first two terms we have:

$\lim_{k\to+\infty}\left[ (1/0!)^k + (1/1!)^k+\sum_{n=2}^{+\infty}\left(\frac{1}{n!}\right )^k\right ]$

equivalently:

$(1/0!)^k + (1/1!)^k+\sum_{n=2}^{+\infty}\lim_{k\to+\infty} \left(\frac{1}{n!}\right)^k$

so:

$\lim_{k\to+\infty}\sum_{n=0}^{+\infty}\left(\frac{ 1}{n!}\right)^k=2$

but this contradicts your tesis that for $\lambda\in]-2;+2[$ the series converges to 1.

I will be satisfied even if it is possible to find an analytical solution for the particular case lambda=1.

4. ## Re: Convergence value of sum_n=0:Inf ( x^n / n! )^k

Oh sorry, I messed up the terms.

New try:
1/1 + lambda^k/1 + lambda^(2k)/2^k + ...

This diverges for lambda>1. Lambda=1 gives a limit of 2, |lambda|<1 gives a limit of 1.
lambda = -1 gives divergence, lambda < -1 would need more analysis.