Let $t\geq 1$ be an integer number, and $p$ a prime number. What fraction can we set on the right side to have a correct inequality? $$\sum_{t\geq 1}\frac{t}{p^{t}-1}< \ldots$$ Thank you for any help.
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2What did you try ? – TheSilverDoe May 03 '23 at 19:04
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I don't know where to start? – Mary May 03 '23 at 19:06
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I don't think this is true - should the inequality be the other way round? – A. Goodier May 03 '23 at 19:09
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2You might start by evaluating the two sides numerically for some small primes $p$. – Robert Israel May 03 '23 at 19:10
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For $p$ a prime, $p>2$, we have $\frac{1}{p-1} > \frac{p}{(p-1)^3}$ and if you take the first couple of terms of the series, you will also find it is false for $p=2$. – A. Goodier May 03 '23 at 19:13
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1The $t=1$ term on the left is already greater than the right side if $p \ge 3$. – Robert Israel May 03 '23 at 19:15
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If $p$ is prime and $t\in\mathbb{N}$, then $\frac{t}{p^t-1} \leqslant \frac{t}{p^{t-1}}$ since $p^t$ and $p^{t-1}$ differ by at least $1$. If you're trying to show the series converges, you can find many questions on the site about how to sum the series $\sum\frac{n}{p^n}$ . – A. Goodier May 03 '23 at 19:32
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@ A. Goodier, Thank you very much. I want to replace a fraction (according to $p$) on the right side to have an true inequality. – Mary May 03 '23 at 19:37
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2Yes, so if you sum the series $\sum\frac{t}{p^{t-1}}$, you will get $\frac{p^2}{(p-1)^2}$ . – A. Goodier May 03 '23 at 19:45
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Copying the answer I gave in the comments:
If $p$ is prime and $t\in\mathbb{N}$, we have $p^{t-1}\leqslant p^t-1$, so $\dfrac{1}{p^t-1}\leqslant \dfrac{1}{p^{t-1}}$ .
Hence $$\sum_{t\ge1}\frac{t}{p^t-1}\le\sum_{t\ge1}\frac{t}{p^{t-1}}=\sum_{t\ge0}\frac{t+1}{p^t}=\sum_{t\ge0}\frac{t}{p^t}+\sum_{t\ge0}\frac{1}{p^t}=\frac{p}{(p-1)^2}+\frac{p}{p-1}=\frac{p^2}{(p-1)^2}\,.$$ To evaluate the last two series, see here for the first series and use the geometric series formula for the second series.
A. Goodier
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@ A. Goodier, Thanks a lot. I saw the previous (incorrect) inequality in a paper published in a math journal!!! Thank you for your help to correct it. Is it possible to replace $\frac{p}{(p-1)^{2}}$ instead of $\frac{p^{2}}{(p-1)^{2}}$ in your answer? – Mary May 03 '23 at 20:00
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@ A. Goodier, unfortunately, in that paper it is assumed the inequality $\sum_{t\geq 1}\frac{t}{p^{t}-1}<\frac{p}{(p-1)^{2}}$ is true! – Mary May 03 '23 at 20:09
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@Mary it must be wrong - you can check numerically that this is false for various values of $p$ – A. Goodier May 03 '23 at 20:11
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