Let  $$a_n = n + n^{-1}$$​. Which of the following is true for the series:
​$$\sum_{n=1}^\infty \frac{(-1)^{n+1} a_{n+1}}{n!}$$ ?

Correct option is D

Given:

$$a_n = n + \frac{1}{n}.$$​​

Define the series:

​$$\sum_{n=1}^\infty \frac{(-1)^{n+1} a_{n+1}}{n!}.$$​​

Let:
​$$b_n = \frac{(-1)^{n+1} a_{n+1}}{n!}.$$​​

Analyze  $$a_{n+1}$$​​
For  $$a_{n+1} :$$​​

​$$a_{n+1} = (n+1) + \frac{1}{n+1}.$$​​

Thus:

​$$\frac{a_{n+1}}{n!} = \frac{n+1}{n!} + \frac{1}{n! (n+1)}.$$​​

Now, rewrite:$$\frac{a_{n+1}}{n!} = \frac{1}{(n-1)!}+\frac{1}{n!} + \frac{1}{n! (n+1)}.$$​​

​$$\textbf{ Check for Convergence}\\$$​​
Take the limit as   $$n \to \infty :$$​​

​$$\lim_{n \to \infty} \frac{a_{n+1}}{n!} = \lim_{n \to \infty} \left( \frac{1}{(n-1)!}+\frac{1}{n!} + \frac{1}{n! (n+1)} \right).$$​​


Since factorial growth dominates any polynomial or rational term:

​$$\lim_{n \to \infty} \frac{a_{n+1}}{n!} = 0.$$​​


By the alternating series test (Leibniz test), the series converges.

​$$\textbf{Evaluate the Series}\\$$​​
Using the expansion of  $$e^x$$​ :

​$$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}.$$​​


​$$\implies e^{-1}=1-1+\frac{1}{2!}-\frac{1}{3!}+\cdots$$​​


Write the given series as:

​$$\sum_{n=1}^\infty \frac{(-1)^{n+1} a_{n+1}}{n!} = \sum_{n=1}^\infty \frac{(-1)^{n+1} [(n+1)^{-1}+(n+1)]}{n!}$$​


​$$= \sum_{n=1}^{\infty}\big (-1)^{n+1}[ \frac{1}{(n-1)!}+\frac{1}{n!}+\frac{1}{(n+1)!}\big]$$​​


​$$=\left(1-1+\frac{1}{2!}-\frac{1}{3!}+\cdots\right)+\left(1-\frac{1}{2!}+\frac{1}{3!}-\cdots\right)+\left(\frac{1}{2!}-\frac{1}{3!}+\cdots\right)$$​​


Adding these, the total is:

​$$e^{-1} + (1 - e^{-1}) + e^{-1} = e^{-1} + 1.$$​​



Conclusion:
The series converges to $$e^{-1} + 1 .$$​​

The correct option is:

​$$\boxed{\text{Option (D)}}.$$​​