1
$\begingroup$

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!

$\endgroup$
3
$\begingroup$

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 $\epsilon >1$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.