Mathematics Asked by alans on December 9, 2020

How to deduce if series is convergent (depending on parameter $a$):

$$

sum_{n=1}^inftyleft[left(1+frac{1}{n}right)^{n+1}-aright]sin{n}?

$$

If $a=e,$ we have that $lim_{ntoinfty}a_n=0,$ but it is not sufficient to show that series converges.

I’m stuck with this today, not at my best. Any help is welcome. Thanks in advance.

If $a neq e$, then $$left[left(1+frac{1}{n}right)^{n+1}-aright]sin{n}$$ does not tend to $0$ so the series cannot converge.

If $a=e$, you have $$left[left(1+frac{1}{n}right)^{n+1}-eright]sin{n} = left[e^{(n+1)lnleft( 1 + frac{1}{n}right)}-eright]sin{n}=eleft[e^{(n+1)lnleft( 1 + frac{1}{n}right)-1}-1right]sin{n}$$

Now $$(n+1)lnleft( 1 + frac{1}{n}right)-1 = (n+1)left( frac{1}{n} - frac{1}{2n^2} + frac{1}{3n^3} + oleft( frac{1}{n^3}right)right) - 1$$ $$= frac{1}{2n} - frac{1}{6n^2} + oleft( frac{1}{n^2}right)$$

So $$left[left(1+frac{1}{n}right)^{n+1}-eright]sin{n} =eleft[ frac{1}{2n} - frac{1}{24n^2} + oleft( frac{1}{n^2}right)right]sin{n}$$

Finally, the series $$sum frac{sin(n)}{n}$$ converges (by Dirichlet's test) ; and the series $$sum frac{sin(n)}{n^2}$$ is absolutely convergent. So your series converges.

Correct answer by TheSilverDoe on December 9, 2020

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