Is the series $\sum_{n=1}^{\infty} \Bigl(1-\Bigl(1-\frac{1}{n^{1+\epsilon}}\Bigr)^n\Bigr)$ convergent?

Question

While i was solving a problem in probability theory I came across the following series $$\sum_{n=1}^{\infty} \biggl(1-\biggl(1-\frac{1}{n^{1+\epsilon}}\biggr)^n\biggr)$$

and in order to complete my solution I want to show the above series converges but I couldn't prove it (I still dont know if it converges).

I tried to go with Taylor expansion of $x \mapsto \ln(1-x^{1+\epsilon})$ but I couldn't get anything interesting.

Then I showed that $1-\bigl(1-\frac{1}{n^{1+\epsilon}}\bigr)^n \leq \frac{1}{n^\epsilon}$ but this doesn't help either.

Let me know if you have any idea!

Answer 1

For large $n$, this term is approximately $n^{-\epsilon}$. The convergence condition is $\epsilon>1$.

Answer 2

We have that

$$\biggl(1-\frac{1}{n^{1+\epsilon}}\biggr)^n=e^{n\log\biggl(1-\frac{1}{n^{1+\epsilon}}\biggr)}=e^{-\frac{1}{n^{\epsilon}}+o\left(\frac{1}{n^{\epsilon}}\right)}=1-\frac{1}{n^{\epsilon}}+o\left(\frac{1}{n^{\epsilon}}\right)$$

thus

$$\biggl(1-\biggl(1-\frac{1}{n^{1+\epsilon}}\biggr)^n\biggr)=\frac{1}{n^{\epsilon}}+o\left(\frac{1}{n^{\epsilon}}\right)$$

which converges for $\epsilon >1$.

Answer 3

You can get explicit constants rather than big or little oh like this:

If $0 < x < 1$, $-\ln(1-x) =\sum_{k=1}^{\infty} \dfrac{x^k}{k} \gt x$ and,

$\begin{array}\\ -\ln(1-x) &=\sum_{k=1}^{\infty} \dfrac{x^k}{k}\\ &=x+\sum_{k=2}^{\infty} \dfrac{x^k}{k}\\ &\lt x+\sum_{k=2}^{\infty} \dfrac{x^k}{2}\\ &\lt x+\dfrac{x^2}{2(1-x)}\\ \end{array} $

Therefore, if $0 < x < \frac12$, $-\ln(1-x) \lt x+x^2 $ so $-x-x^2 \lt\ln(1-x) \lt -x $.

Similarly, if $0 < x < 1$, $\exp(x) =\sum_{k=0}^{\infty} \dfrac{x^k}{k!} \gt 1+x$ and $\exp(x) \lt\sum_{k=0}^{\infty} x^k =\dfrac1{1-x} $.

Therefore, for $0 < x < 1$, $1-x \lt \exp(-x) \lt \dfrac1{1+x}$.

Therefore, if $n^{-c} < 1$, then

$\begin{array}\\ (1-\frac1{n^{1+c}})^n &=\exp(n\ln(1-\frac1{n^{1+c}}))\\ &\lt\exp(n(-\frac1{n^{1+c}}))\\ &=\exp(-n^{-c})\\ &\lt \dfrac1{1+n^{-c}}\\ \end{array} $

so

$\begin{array}\\ 1-(1-\frac1{n^{1+c}})^n &\gt 1-\dfrac1{1+n^{-c}}\\ &=\dfrac{n^{-c}}{1+n^{-c}}\\ &=\frac12 n^{-c}\\ \end{array} $

so the sum diverges if $c \le 1$.

If $c > 1$ then

$\begin{array}\\ (1-\frac1{n^{1+c}})^n &=\exp(n\ln(1-\frac1{n^{1+c}}))\\ &\gt\exp(n(-\frac1{n^{1+c}}-\frac1{n^{2+2c}}))\\ &=\exp(-n^{-c}-n^{-1-2c})\\ &\gt 1-(n^{-c}+n^{-1-2c})\\ \end{array} $

so

$\begin{array}\\ 1-(1-\frac1{n^{1+c}})^n &\lt 1-(1-(n^{-c}+n^{-1-2c}))\\ &= n^{-c}+n^{-1-2c}\\ &\lt n^{-c}+n^{-3}\\ \end{array} $

and the sum converges.