MCQ
$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\frac{1}{n}{e^{\frac{r}{n}}}} $ is
- A$e + 1$
- ✓$e - 1$
- C$1 - e$
- D$e$
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Then $\lim _{a \rightarrow \infty} \frac{a(a+1)}{2} \tan ^{-1}\left(\frac{1}{a}\right)+a^2-2 \log _e a$ is equal to