Question

Theory and ExamplesSuppose that a₁, a₂, a₃, …, aₙ are positive numbers satisfying the following conditions:i) a₁ ≥ a₂ ≥ a₃ ≥  …;ii) the series a₂ + a₄ + a₈ + a₁₆ + … diverges.Show that the seriesa₁/1 + a₂/2 + a₃/3 + …diverges.

Accepted Answer

First, carefully analyze the given conditions: the sequence $$ a_1, a_2, a_3, \ldots, a_n $$ consists of positive terms and is non-increasing, i.e., $$ a_1 \geq a_2 \geq a_3 \geq \ldots . $$This monotonicity will be crucial in comparing terms. Next, observe the series $$ a_2 + a_4 + a_8 + a_{16} + \ldots $$ which is given to diverge. Notice that the indices are powers of 2, so this is a subseries of the original sequence taken at exponentially growing indices. To connect the divergence of the subseries to the original series $$ \sum_{n=1}^\infty \frac{a_n}{n} $$, group the terms of the original series into blocks corresponding to intervals between powers of 2. For example, consider the blocks $$ [2^k, 2^{k+1} - 1] $$ for $$ k = 0, 1, 2, \ldots . $$Within each block $$ [2^k, 2^{k+1} - 1] $$, use the fact that $$ a_n  is $$non-increasing to bound $$ a_n $$ from below by $$ a_{2^{k+1}}  (or a $$similar term). Then, estimate the sum of $$ \frac{a_n}{n} $$ over this block by comparing it to $$ a_{2^{k+1}} $$ times the sum of $$ \frac{1}{n} $$ over the block. Finally, use the divergence of the subseries $$ a_2 + a_4 + a_8 + \ldots $$ and the harmonic-like growth of the sums $$ \sum_{n=2^k}^{2^{k+1}-1} \frac{1}{n}  to $$conclude that the original series $$ \sum_{n=1}^\infty \frac{a_n}{n} $$ must also diverge.

Theory and Examples
Suppose that a₁, a₂, a₃, …, aₙ are positive numbers satisfying the following conditions:
i) a₁ ≥ a₂ ≥ a₃ ≥ …;

ii) the series a₂ + a₄ + a₈ + a₁₆ + … diverges.
Show that the series

a₁/1 + a₂/2 + a₃/3 + …

diverges.

Verified video answer for a similar problem:

Key Concepts

Monotone Sequences

Divergence of Subseries

Comparison Test for Series