Question

Which one of the following series is convergent?

• A. $\displaystyle \sum_{n=1}^{\infty} \left( \frac{5n + 1}{4n + 1} \right)^n$
• B. $\displaystyle \sum_{n=1}^{\infty} \left(1 - \frac{1}{n}\right)^n$
• C. $\displaystyle \sum_{n=1}^{\infty} \frac{\sin n}{n^{1/n}}$
• D. $\displaystyle \sum_{n=1}^{\infty} \sqrt{n}\left(1 - \cos\left(\frac{1}{n}\right)\right)$

Hint:

For convergence, always check term behavior for large $n$; use approximations like $\cos x \approx 1 - \frac{x^2}{2}$ for small $x$.

Answer 1

The Correct Option is D

Step 1: Analyze each option.
(A) $\left(\frac{5n+1}{4n+1}\right)^n \approx \left(\frac{5}{4}\right)^n$ which diverges since ratio > 1.
(B) $\left(1 - \frac{1}{n}\right)^n \to \frac{1}{e}$, so terms do not approach 0; hence the series diverges.
(C) $\frac{\sin n}{n^{1/n}}$: Since $n^{1/n} \to 1$, terms behave like $\sin n$ which do not tend to 0; hence diverges.
(D) For small $x$, $1 - \cos x \approx \frac{x^2}{2}$, thus \[ \sqrt{n}\left(1 - \cos\left(\frac{1}{n}\right)\right) \approx \sqrt{n}\cdot \frac{1}{2n^2} = \frac{1}{2n^{3/2}}. \] The series $\sum \frac{1}{n^{3/2}}$ converges (p-series, $p>1$).

Step 2: Conclusion.
Hence, the only convergent series is (D).