Question:
Which of the following series is divergent?
Which of the following series is divergent?
Show Hint
If $\sin(1/n)$ appears, always compare with $1/n$.
Updated On: Jun 29, 2026
- $\sum \sin\left(\frac{1}{n}\right)$
- $\sum \left(1+\frac{1}{\sqrt{n}}\right)^{-n^{1/2}}$
- $\sum \left(1+\frac{1}{n}\right)^{-n^{2}}$
- $\sum \frac{1}{(\log n)^n}$
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
A series diverges if its general term does not go to zero fast enough or behaves like a divergent comparison series.
Step 1: Analyze option (A).
For small $x$: \[ \sin x \approx x \] So: \[ \sin\left(\frac{1}{n}\right)\approx \frac{1}{n} \]
Step 2: Compare with harmonic series.
\[ \sum \frac{1}{n} \text{ diverges} \] So (A) diverges.
Step 3: Check other options briefly.
(B), (C), (D) all behave like rapidly decaying exponential-type series → convergent. \[ \Rightarrow \text{Only (A) diverges} \]
Step 1: Analyze option (A).
For small $x$: \[ \sin x \approx x \] So: \[ \sin\left(\frac{1}{n}\right)\approx \frac{1}{n} \]
Step 2: Compare with harmonic series.
\[ \sum \frac{1}{n} \text{ diverges} \] So (A) diverges.
Step 3: Check other options briefly.
(B), (C), (D) all behave like rapidly decaying exponential-type series → convergent. \[ \Rightarrow \text{Only (A) diverges} \]
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