Series Convergence $\sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{2n+2}{2n+4}\right)^n$ I have to show if this series converges or diverges. I tried using asymptotics, but it's not formally correct as they should work only when the arguments are extremely small. Any ideas?
$$
\sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{2n+2}{2n+4}\right)^n
$$
 A: Recall that, as $x \to 0$, by the Taylor expansion, we have
$$
\begin{align}
e^x& =1+x+\mathcal{O}(x^2)\\
\ln (1+x)&=x-\frac {x^2}{2}+\mathcal{O}(x^3)
\end{align}
$$ giving, as $n \to \infty$,
$$
n\ln \left(1-\frac {2}{2n+4}\right)=n \left(-\frac {2}{2n+4}+\mathcal{O}(\frac {1}{n^2})\right)=-1+\mathcal{O}(\frac {1}{n})
$$ and
$$
\begin{align}
\frac{1}{n} \left(\frac{2n+2}{2n+4}\right)^n&=\frac{1}{n} \left(\frac{2n+4-2}{2n+4}\right)^n\\\\
&=\frac{1}{n} \left(1-\frac {2}{2n+4}\right)^n\\\\
&=\frac{1}{n}e^{n\ln (1-\frac {2}{2n+4})}\\\\
&=\frac{1}{n}e^{-1+\mathcal{O}(\frac {1}{n})}\\\\
&\sim \frac{e^{-1}}{n}
\end{align}
$$ thus your initial series is divergent.
A: Hint: Put $n=k-2$ and simplify the fraction inside the parentheses.
A: $\frac{2n+2}{2n+4} = \frac{n+1}{n+2}$
With $\frac{n+1}{n+2} \le  1+\frac1n$ and $\lim_{n\to \infty} (1+\frac1n)^n = e$ you can use comparison test to show that it is divergent.
A: Without any additional algebra, just notice that after cancellations both expressions
$$
\frac{\bigg(1+\frac{1}{2n}\bigg)^n}{\bigg(1+\frac{2}{n}\bigg)^n} = \frac{O(1)}{O(1)}=O(1)
$$
Hence your summand is $f_n = \frac{1}{n} \cdot O(1)$.
