← 2015 Paper 2

UPSC 2015 Maths Optional Paper 2 Q1c — Step-by-Step Solution

10 marks · Section A

Absolute and conditional convergence · Real Analysis · asked 4× in 13 yrs · Read the full method →

Question

Test the convergence and absolute convergence of the series $\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{n}{n^2+1}$.

Technique

Leibniz test for alternating convergence; limit comparison with $\sum 1/n$ for the absolute value series.

Solution

Let $a_n=\dfrac{n}{n^2+1}$. The series is $\sum(-1)^{n+1}a_n$.

Step 1 — Convergence via Leibniz’s alternating series test

Conditions:

1. $a_n>0$ — yes (positive for $n\ge 1$).
2. $a_n\to 0$ — $a_n=n/(n^2+1)\sim 1/n\to 0$. ✓
3. $a_n$ eventually decreasing — show this.

Monotonicity. Let $f(x)=x/(x^2+1)$. Then

For $x>1$, $f'(x)<0$. So $\{a_n\}$ is strictly decreasing for $n\ge 2$ (and $a_1=1/2$, $a_2=2/5=0.4$, so $a_1>a_2$ also).

All three conditions met. By Leibniz, the series converges.

Step 2 — Absolute convergence — test $\sum|a_n|=\sum\dfrac{n}{n^2+1}$

For large $n$: $\dfrac{n}{n^2+1}\sim\dfrac{1}{n}$. Compare with the harmonic series $\sum 1/n$, which diverges.

Limit comparison test. $\displaystyle\lim_{n\to\infty}\dfrac{n/(n^2+1)}{1/n}=\lim_{n\to\infty}\dfrac{n^2}{n^2+1}=1$.

Limit is positive and finite, so $\sum n/(n^2+1)$ and $\sum 1/n$ either both converge or both diverge. Since $\sum 1/n$ diverges, so does $\sum|a_n|$.

Therefore $\sum(-1)^{n+1}a_n$ does not converge absolutely.

Step 3 — Conclusion

Answer