← 2016 Paper 2

UPSC 2016 Maths Optional Paper 2 Q2a — Step-by-Step Solution

15 marks · Section A

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

Question

Show that the series $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n+1}$ is conditionally convergent. (If you use any theorem(s) to show it, then you must give a proof of that theorem(s).)

Technique

Leibniz test (proved via monotone bounded even partial sums) for convergence; harmonic divergence (proved by dyadic blocking) for non-absolute-convergence.

Solution

“Conditionally convergent” means: the series converges, but the series of absolute values diverges. We prove convergence by the Leibniz alternating-series test (with proof) and divergence of the absolute series by comparison with the harmonic series (with proof).

Step 1 — Convergence: Leibniz alternating series test (statement + proof)

Write the series as $\sum(-1)^{n+1}a_n$ with $a_n=\dfrac1{n+1}>0$.

Leibniz Test. If $(a_n)$ is positive, monotonically decreasing, and $a_n\to0$, then $\sum_{n=1}^\infty(-1)^{n+1}a_n$ converges.

Proof. Let $S_N=\sum_{n=1}^N(-1)^{n+1}a_n$. Consider even partial sums:

Each bracket is $\ge0$ (since $a_n$ decreasing), so $(S_{2m})$ is non-decreasing. Also

so $(S_{2m})$ is bounded above by $a_1$. By the Monotone Convergence Theorem $S_{2m}\to S$ for some $S$. For odd sums, $S_{2m+1}=S_{2m}+a_{2m+1}$, and $a_{2m+1}\to0$, so $S_{2m+1}\to S$ too. Even and odd subsequences share the limit $S$, hence $S_N\to S$. $\square$

Apply it. $a_n=\dfrac1{n+1}$: positive; decreasing since $a_{n+1}=\dfrac1{n+2}<\dfrac1{n+1}=a_n$; and $a_n=\dfrac1{n+1}\to0$. All hypotheses hold, so

Step 2 — The absolute series diverges (comparison with the harmonic series, with proof)

The series of absolute values is

which is the harmonic series with its first term omitted.

Claim. $\sum_{n=1}^\infty\dfrac1{n+1}$ diverges.

Proof (Cauchy condensation / grouping). It suffices to show $\sum_{k=2}^\infty\frac1k=\infty$ (dropping the finite first term $1$ of the harmonic series does not change divergence). Group the harmonic series in blocks of lengths $1,2,4,8,\ldots$:

The block from index $2^{j}+1$ to $2^{j+1}$ has $2^{j}$ terms each $\ge\dfrac1{2^{j+1}}$, so its sum is $\ge 2^{j}\cdot\dfrac1{2^{j+1}}=\dfrac12$. Hence the partial sums exceed $\dfrac12\cdot(\text{number of completed blocks})\to\infty$. So the harmonic series, and therefore $\sum\frac1{n+1}$, diverges. $\square$

Step 3 — Conclusion

The series converges (Step 1) but does not converge absolutely (Step 2). Therefore

Answer