Vtyjkyuk

i need some advice on how to prove if this series converges or diverges
tried using D'Alembert test but i received $1$ so it does not help me.
Thanks.

$$sum_n=1^infty fracn!e^nn^n$$

The series converges and we can see that in so many ways that is

• root test (maybe the simpler)


• ratio test (also effective)


• limit comparison test (for example with $sum frac1n^2$)


• etc.

As an alternative, note that eventually, that is for $nge e^2$ we have $$n^nge e^2n$$ and since eventually $e^n ge n^3$

$$fracne^nn^nlefracne^ne^2n=frac ne^nle fracnn^3=frac1n^2$$

the given series converges by direct comparison test with $sum frac1n^2$.

Hint: The $n$th term equals

$$eleft(fracenright )^n-1$$

and $e/n le e/3$ for $n>2.$

With some simple manipulations you can show that you can omit the factor $n$ in the numerator (are you able to do this?

Then you are left with $a_n=frace^nn^n$. To see that this sum converges it is enough to see, that these summands are smaller than lets say $left(frace10right)^n$, that is they are smaller than the summands of a (converging) geometric sum.

With ratio test:
$$fraca_n+1a_n=$$
$$fracn+1n frace^n+1e^n fracn^n(n+1)^n+1=$$
$$efracn+1nfracn^n(n+1)^n+1=$$
$$efracn+1nfrac1n+1left(fracnn+1right)^n=$$
$$fracenleft(fracnn+1right)^n to 0$$
Because $left(fracnn+1right)^n to frac1e$ and $fracen to 0$