Root test for $\sum_{n=8}^\infty \left(1+\frac{3}{n}\right)^{n^2}$

Question

I got that it would be infinity and diverge, but my answer seems to be incorrect. What did I do wrong?

$$\lim_{n\to\infty}\left(1+\frac3n\right)^{n^2}= \lim_{n\to\infty} \left(1+\frac3n\right)^{n}= \infty$$

Answer 1

$$L=\lim_{x\rightarrow \infty }(1+\frac{3}{n})^{n^2/n}=\lim_{x\rightarrow \infty }(1+\frac{3}{n})^n$$ after that take the log for both sides $$\log(L)=n\log(\lim_{x\rightarrow \infty }(1+\frac{3}{n}))$$ by using the Lopital Rule the limit is $e^3>1$

so the series is diverge

Answer 2

The limit is of the form:

$$\lim_{n\to\infty}\left(1+\frac{m}{x}\right)^{nx}~\textrm{with }m=3,n=1$$

We know that the limits in these form have the solution $e^{mn}$. Refer to this link for all the limit-related identities.

For the problem here, the limit is $e^3$.

Answer 3

$$\lim_{n \to \infty}\left(1+\frac{3}{n}\right)^n=e^3$$ Moreover the root test is applied on the $n$th term. You don't find the sum of the series like that.The test is only for convergence or divergence.

Answer 4

There is one more great mistake: The limit in the root test does not equal the sum of the series!!