For which values the following summation converges

Question

I have the following problem:

For which $a$ and $b$ values does the following summation converges:

$$\sum_{n=1}^{\infty}(-1)^{n-1}\left(a-\frac{(n-b)^n}{n^n}\right)$$

I tried to solve this in many ways but keep failing since I don't know how to deal with this kind of questions especially with two parameters and not one.

Answer 1

$a\ne 1$ the series terms $\to \pm (a-1)$ which won't converge.

For $a=1$, two cases:

(1) $b\ne 0$, the series is alternately decreasing, so it converges (conditionally).

(2) $b=0$ all terms $=0$, so convergence is trivial.

Answer 2

hint

$$\frac{(n-b)^n}{n^n}=(1-\frac bn)^n$$

$$=e^{n\ln(1-\frac bn)}$$

$$e^{-b}(1-\frac{b^2}{2n})+\frac 1n\epsilon(n))$$ If the series converges, necessarily, its general term must go to zero. thus, we must have $$a=e^{-b}$$

Answer 3

Let $a_n=a-\left(1-\frac{b}{n}\right)^n$, then the series is $\sum_{n=1}^\infty (-1)^{n-1}a_n$.

• You will need to have $\lim_{n\to \infty}a_n=0\iff \lim_{n\to\infty}\left(1-\frac{b}{n}\right)^n=e^{-b}=a$

• Also try to show that for each $b\in \mathbb{R},$ $a_n>0$ and $(a_n)$ decreases monotonically from some index $N$, then convergence follows from the alternating series test.