I need to find out for what parameters a,b converge/diverge this sum.I tried integral test and i know that it converge for a>1 but i do not know how to continue.Thanks in advance for advice.
$$\sum _{n=1}^{\infty }\frac{1}{n^aln^b(n)}$$
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Hint: $\ln n = o(n^\epsilon)$ for any $\epsilon > 0$ however small, which proves divergence if $a < 1$. – Connor Harris Aug 26 '19 at 18:56
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Are you sure the sum starts at $1$? The reason I ask is because if we have a $\ln(1)$ in the denominator, then the first term is undefined. – Michael Aug 26 '19 at 19:03
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@ConnorHarris That is not true for all values of b(Take for example b=0) – miraunpajaro Aug 26 '19 at 19:03
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@miraunpajaro Huh? If $b = 0$, then the sum is just $\sum_{n=1}^\infty \frac{1}{n^a}$, which also diverges if $a < 1$. – Connor Harris Aug 26 '19 at 19:06
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@Connor Harris My bad. I thought your comment said convergence. I somehow managed to misread it – miraunpajaro Aug 26 '19 at 19:10
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It would probably be a good idea to start the sum at $n=2$ instead of $n=1$, since otherwise you have a power of $\ln(1)=0$ in the denominator. – Barry Cipra Aug 26 '19 at 20:41
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This is a special case of the Generalized harmonic series of type 2 (found only italian link)
Which is :
$$\sum\limits_{n=1}^{+\infty} \frac{1}{n^{\alpha}\log^{\beta}(n)} \begin{cases} \text{converges }\hspace{0.2cm} \alpha > 1 \vee \alpha = 1 \wedge \beta > 1 \\ +\infty \hspace{0.2cm} \alpha < 1 \vee \alpha = 1 \wedge \beta \leq 1 \end{cases}$$
Typically you reach this conclusion using Cauhcy condensation test and asymptotic comparison with the generalized harmonic series $\sum\limits_{n=1}^{+\infty} \frac{1}{n^{\alpha}}$
Bernard
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jacopoburelli
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If you read French, they're known in France as Bertrand's series. There are also Spanish an Romanian versions. – Bernard Aug 26 '19 at 19:49
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Thanks! I really didn't know that, should i re-edit it ? @Bernard – jacopoburelli Aug 26 '19 at 20:44