Testing convergence of $\sum\limits_{n=1}^{\infty}u_n$ , where $u_n = \left ( 4- \frac{1}{n}\right) ^ { \frac{( - 1) ^ {n }}{ n}}$

Question

What I did was: I tested for $\lim_\limits{n\to\infty}u_n$ by taking log

$$\lim_\limits{n\to\infty} \frac{\ln\ \left(4 - \frac{1}{n}\right)} {\frac{n}{(-1)^n}}$$

Applying L'hopital's rule,

$$\lim_\limits{n\to\infty} \frac{\left(\frac{1} {4-\frac{1}{n}}\right)\left(\frac{1}{n^2}\right)}{(-1)^n-n^2(-1)^{n-1}}$$

The denominator in above value will either be $1+n^2$ or $-(1+n^2)$

$$\lim_\limits{n\to\infty} \frac{\left(\frac{1} {4-\frac{1}{n}}\right)\left(\frac{1}{n^2}\right)}{1+n^2}$$

$$=\frac{1}{(4n-1)(n)(1+n^2)}\\$$

On applying the limit we get $\frac{1}{\infty}$, which is $0$

But the actual answer is $1$, which would also prove that series diverges.

Where am I going wrong?

Answer 1

L'hopital's Rule can only be applied in certain cases where the given limit takes an indeterminate form like $\frac {0} {0}$ or $\frac {\infty} {\infty}$. You don't have such a form here and you can get the limit by just taking the limit of the numerator and the denominator. In fact $\lim u_n = 4^{0}=1$ from directly from teh given expression for $u_n$.

Answer 2

If you have proven that $\lim_\limits{n \to \infty}\ln u_n = 0$, which implies that$\lim_\limits{n \to \infty} u_n = 1$, hence the series diverges.

Remark and credit: Kavi is right that L'hopital's rule is not applicable here.

$$\lim\limits_{n \to \infty} |\ln u_n |\le \lim\limits_{n \to \infty} \frac{\ln 4}{n}=0$$

Hence $\lim_{n \to \infty} \ln u_n =0$, and $\lim_\limits{n \to \infty} u_n = 1$