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Let $0<a_n<1$ be real numbers. Prove that $\sum \frac{a_n}{\ln a_n}$ converges $\implies \sum \frac{a_n}{\ln (1+n)}$ converges

Note that $\displaystyle \frac{a_n}{\ln (1+n)} \leq \frac{a_n}{-\ln a_n}\iff a_n\geq \frac{1}{n+1}$

If $a_n\geq \frac{1}{n+1}$ for sufficiently many $n$, the series $\sum \frac{a_n}{\ln (1+n)}$ can thefore be compared with $\sum \frac{a_n}{-\ln a_n}$ and we're done. Otherwise, $a_n< \frac{1}{n+1}$ and we're tempted to consider the series $\displaystyle \sum \frac{1}{(n+1)\ln (1+n)}$ which is unfortunately divergent.

That's as I far as I can go with this problem. Summation by part is a dead end as well.

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Edit/Remark: this is very similar in spirit to mathguy's answer below/above; only a lot more verbose.


We will start from the "trivial" equality $$ \frac{a_n}{\ln(n+1)} = \frac{a_n}{\ln(n+1)} \mathbb{1}_{\left\{ a_n < \tau_n\right\}} + \frac{a_n}{\ln(n+1)} \mathbb{1}_{\left\{ a_n \geq \tau_n \right\}} $$ which holds for any choice of "threshold" $(\tau_n)_n$; specifically, we want to choose a suitable sequence of thresholds which would enable us to invoke theorems of comparison. (In what you did, you implicitly chose $\tau_n = \frac{1}{n+1}$: this is not enough, so we'll choose a smaller threshold.)

Moreover, to make $\ln(n+1)$ appear somewhere (namely, when upper bounding the second term of the above identity), we will want to choose $\tau_n=\frac{1}{(n+1)^\alpha}$ for some $\alpha > 0$: so that $\ln\frac{1}{\tau_n} = \alpha\ln(n+1).$

(Stop now if you didn't want the full solution, but only a hint.)


In detail: As we'll see, any $\alpha > 1$ will do -- below, we set $\alpha \stackrel{\rm def}{=}2$. Write $$ \frac{a_n}{\ln(n+1)} = \frac{a_n}{\ln(n+1)} \mathbb{1}_{\left\{ a_n < \frac{1}{(n+1)^2} \right\}} + \frac{a_n}{\ln(n+1)} \mathbb{1}_{\left\{ a_n \geq \frac{1}{(n+1)^2} \right\}} $$ For the first term, we can write $$\frac{a_n}{\ln(n+1)} \mathbb{1}_{\left\{ a_n < \frac{1}{(n+1)^2} \right\}}\leq \frac{1}{(n+1)^2\ln(n+1)} \mathbb{1}_{\left\{ a_n < \frac{1}{(n+1)^2} \right\}} \leq \frac{1}{(n+1)^2\ln(n+1)}$$

For the second, $$\begin{align} \frac{a_n}{\ln(n+1)} \mathbb{1}_{\left\{ a_n \geq \frac{1}{(n+1)^2} \right\}} &= \frac{a_n}{\frac{1}{2}\ln((n+1)^2)} \mathbb{1}_{\left\{ a_n \geq \frac{1}{(n+1)^2} \right\}} \leq \frac{a_n}{\frac{1}{2}\ln \frac{1}{a_n}} \mathbb{1}_{\left\{ a_n \geq \frac{1}{(n+1)^2} \right\}} \\ &\leq \frac{2a_n}{\ln \frac{1}{a_n}} \mathbb{1}_{\left\{ a_n \geq \frac{1}{(n+1)^2} \right\}} \leq \frac{2a_n}{\ln \frac{1}{a_n}}. \end{align}$$

Together, we get $$ 0 < \frac{a_n}{\ln(n+1)} \leq \frac{1}{(n+1)^2\ln(n+1)} + 2\frac{a_n}{\ln \frac{1}{a_n}} $$ which allows you to conclude by theorems of comparison.

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  • $\begingroup$ @mathguy I didn't realize until the end. I've edited my answer to add something pointing this out. $\endgroup$ – Clement C. Apr 21 '16 at 17:22
  • $\begingroup$ Nicely done, thanks for the thorough answer. $\endgroup$ – Gabriel Romon Apr 21 '16 at 17:36
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Great start.

Now imagine you could compare against $\sum \frac 1 {(n+1)^2\ln(n+1)}$... in your approach, compare the terms of the "new" series against 2 times the terms of the "old" series (times -1). This will lead you to the convergent series for the "remaining" terms.

As you put things together, to make the "sufficiently many $n$" rigorous, compare the new series against the SUM of 2 times the old series (times -1) AND the convergent series $\sum \frac 1 {(n+1)^2\ln(n+1)}$.

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