# Summation with two costs

I need to find for which alpha and beta values, the following summation will converge and for which it will diverge

$$\sum\limits_{n=1}^\infty (-1)^{n-1}\left(\alpha-\frac{(n-\beta)^n}{n^n}\right)$$

Any ideas on what law could be helpful with this one? I'm confused with the fact that I have 2 unknown costs and not only one which makes everything much more complex to me.

I'm going to provide a partial answer. If you want that series be convergent you need at least that $$\lim_{n\to \infty } \alpha - \left( 1-\frac{\beta}{n}\right)^{n} =\alpha-e^{-\beta} = 0$$ So $$\beta =-\ln(\alpha)$$, also observe that $$a_n:= \alpha - \left( 1-\frac{\beta}{n}\right)^{n}$$ satisfies $$a_n>a_{n+1}\geq ... \geq \inf_{n \in \mathbb{N}}{a_n}=0$$, when $$\beta$$ is negative, for this you'll need $$\alpha>1$$ (Check what happen when $$\alpha=1$$). So in that case by Leibnitz's Criterion for all $$\alpha>1$$ you'll have that $$\sum_{n=1}^{\infty}(-1)^na_n < \infty.$$
1. Anyway for convergence of $$\sum\limits_{n=1}^\infty (-1)^{n-1}\left(\alpha-\frac{(n-\beta)^n}{n^n}\right)$$ the limit $$\lim\limits_{n\to\infty} (-1)^{n-1}\left(\alpha-\frac{(n-\beta)^n}{n^n}\right)$$ must be $$=0$$, so $$\lim\limits_{n\to\infty} \alpha-\frac{(n-\beta)^n}{n^n}=0.$$ As $$\lim\limits_{n\to\infty} \frac{(n-\beta)^n}{n^n}= \lim\limits_{n\to\infty} (1-\frac{\beta}{n})^n=e^{-\beta}$$ for an arbitrary $$\beta$$, $$\alpha$$ should be $$=e^{-\beta}$$ or the summation diverges.
2. Considering $$\sum\limits_{n=1}^\infty (-1)^{n-1}\left(e^{-\beta}-(1-\frac{\beta}{n})^n\right)$$ it converges by alternating series test for an arbitrary $$\beta$$: it suffices to show that $$\forall n>N$$ $$e^{-\beta}-(1-\frac{\beta}{n+1})^{n+1} $$(1-\frac{\beta}{n+1})^n(1-\frac{\beta}{n+1})>(1-\frac{\beta}{n})^n$$ $$\left(\frac{1-\frac{\beta}{n+1}}{1-\frac{\beta}{n}}\right)^n(1-\frac{\beta}{n+1})>1$$ $$\left(\frac{n^2+n-\beta n}{n^2+n-\beta (n+1)}\right)^n(1-\frac{\beta}{n+1})>1$$ $$\left(1+\frac{\beta}{(n-\beta) (n+1)}\right)^n(1-\frac{\beta}{n+1})>1$$ Then, thanks to this answer, by Bernoulli inequality we have $$\left(1+\frac{\beta}{(n-\beta) (n+1)}\right)^n \geq 1+n\frac{\beta}{(n-\beta)(n+1)}\hbox{, hence}$$ $$\left(1+\frac{\beta}{(n-\beta) (n+1)}\right)^n(1-\frac{\beta}{n+1}) \geq \left(1+\frac{n}{(n-\beta)}\frac{\beta}{(n+1)}\right)(1-\frac{\beta}{n+1})=$$ $$1+\frac{\beta^2}{(n + 1)^2 (n - \beta)}>1$$ for $$n>\beta$$. So
Answer: the summation converges when $$\alpha=e^{-\beta}$$ and diverges otherwise.
• By MA-MG $\sqrt[n+1]{(1-\frac{\beta}{n})^n} \leq \frac{(1-\frac{\beta}{n})+...+(1-\frac{\beta}{n})+1}{n+1} = 1-\frac{\beta}{n+1}$ May 21, 2020 at 3:13