determine whether the following series converge or diverge? determine whether the following series converge  or diverge ?
$a)$  $\sum_{n=1}^{\infty} (\frac {n}{n+1})^{n(n+1)}$
$b)$ $\sum_{n=1}^{\infty} (\frac {n^2 +1 }{n^2 +n +1})^{n^2}$
My attempts : for   $a)$ $(\frac {n}{n+1})^{n(n+1)}= (1- \frac{1}{n+1})^{n(n+1)}= e^{-n}$, now  $\sum_{n=1}^{\infty}\frac{1}{e^n}$ is  converge 
For  $b )$ $(\frac {n^2 +1 }{n^2 +n +1})^{n^2}=( 1-  \frac{n}{n^2+n+1})^{n. n}$...after  that  i can not able to procedd Further,,,,
Any hints/ solution will be appreciated
thanks in advance
 A: By root test we have that
$$\sqrt[n]{a_n}=\left(\frac {n}{n+1}\right)^{n+1}=\left(1-\frac {1}{n+1}\right)^{n+1}\to \frac1e$$
$$\sqrt[n]{b_n}=\left(\frac {n^2 +1 }{n^2 +n +1}\right)^{n}=\left[\left(1-\frac {n}{n^2+n+1}\right)^{\frac{n^2+n+1}{n}}\right]^{\frac{n^2}{n^2+n+1}} \to \frac1e$$
therefore $\sum a_n$ and $\sum b_n$ both converge.
A: You should write $(1- \frac{1}{n+1})^{n(n+1)}\sim e^{-n}$ instead of $(1- \frac{1}{n+1})^{n(n+1)}= e^{-n}.$  The expressions are asymptotic, not equal.  (That is, the limit of their ratio as $n\to\infty$ is $1$.)  Other than that, your calculation in part a) is correct, so far as I can see.
For part b) you've started out correctly.  Note that as $n\to\infty,$ $\frac{n}{n^2+n+1}$ will behave like $\frac{1}{n}.$  So if you prove that $$ 
1-\frac{n}{n^2+n+1}\sim1-\frac{1}{n}
$$
you'll be able to proceed as in part a).
A: Comparison Test. For a) we have $(\frac {n}{n+1})^{n+1}=(1-\frac {1}{n+1})^{n+1}=e^{-1}(1+\delta_n)$ where  $\lim_{n\to \infty}\delta_n=0.$ So for all but finitely many $n$ we have $(\frac {n}{n+1})^{n(n+1)}=(e^{-1}(1+\delta_n))^n<(e^{-1}(2))^n=(\frac{2}{e})^n.$ In fact this holds for all $n\in \Bbb N.$
Ratio Test. For a)   the ratio $t_n/t_{n+1}$ of the $n$th term $t_n$ to the $(n+1)$th term $t_{n+1}$ is $t_n/t_{n+1}=A_n B_n$ where $A_n=\left(\frac {n(n+2)}{(n+1)^2}\right)^{(n+1)^2}=\left(1-\frac {1}{(n+1)^2}\right)^{(n+1)^2}$ and $B_n= \left(\frac {n+2}{n}\right)^{n+1}=\left(1+\frac {2}{n}\right)^{n+1}.$  Observe that $A_n\to e^{-1}$ while $B_n\to e^2.}
For b) we  have $\left(1-\frac {n}{n^2+n+1}\right)^{n^2}=$ $\left((1-\frac {1}{C(n)})^{C(n)}\right)^{D(n)}$ where $C(n)=n+1+\frac {1}{n}$ and $D(n)=\frac {n^2}{C(n)}.$ Now $C(n)\to \infty$ as $n\to \infty,$ and  $D(n)>n/2$ for all but finitely many $n.$ So for all but finitely many $n,$  the $n$th term $T_n,$ of the series, satisfies $T_n <(2/e)^{D(n)}<(2/e)^{n/2}.$ Note that we don't need very precise estimates for $T_n$  to prove  convergence in this Q.
Nothing wrong with the other answers. There are many ways to solve this.
