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My question is: Does the infinite series
$\sum_{n=1}^\infty \frac{1}{n^{\frac{4}{5}}+10^{10}}$ converge or diverge? I know that $\frac{1}{n^{\frac{4}{5}}}$ diverges by the $p$-test, and that adding the constant $10^{10}$ in the denominator doesn't effect its behavior. I just do not know how to go about proving it. Thanks for the help.
EDIT: Also I tried setting up the integral test (as this is assigned in that section) and got
$$\int_1^\infty \frac{1}{x^{\frac{4}{5}}+10^{10}} dx$$ but can't figure out where to go from there either.

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  • $\begingroup$ Sorry can you change $n$ to $x$ under integral. I can not edit just one symbol, that I forgot. :) $\endgroup$ – Caran-d'Ache Apr 16 '13 at 5:31
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Note that $$\dfrac1{n^{4/5} + 10^{10}} \geq \dfrac1{10^{20}+1} \cdot \dfrac1{n^{4/5}} \,\,\,\,\, \forall n \in \mathbb{Z}^+$$Now conclude what you want.

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  • $\begingroup$ ahh so I can use the integral test to show the series you gave diverges, then the original diverges by comparison, sound about right? $\endgroup$ – underbar Apr 16 '13 at 5:30
  • $\begingroup$ @underbar You can either use the integral test or the $p$ test to prove that $\sum \dfrac1{n^{4/5}}$ diverges. $\endgroup$ – user17762 Apr 16 '13 at 5:31
  • $\begingroup$ alright I think I've got it now. thanks for the help everyone! i'd vote up but I don't have enough rep quite yet lol $\endgroup$ – underbar Apr 16 '13 at 5:33
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I guess you have infinite series and so adding or removing finite terms will not affect the convergence. You can change the start of the summation index. So, just set $n^\frac{4}{5}+10^{10}=k^\frac{4}{5}$ and then $$\sum_{n=1}^\infty \frac{1}{n^{\frac{4}{5}}+10^{10}}=\sum_{k=1+10^{10}}^\infty \frac{1}{k^{\frac{4}{5}}}=\sum_{k=1}^\infty \frac{1}{k^{\frac{4}{5}}}-\sum_{k=1}^{10^{10}} \frac{1}{k^{\frac{4}{5}}}=\sum_{k=1}^\infty \frac{1}{k^{\frac{4}{5}}}-\mathrm{some}\ \mathrm{finite} \ \mathrm{numer}$$ Which diverges.

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