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Determine whether the series $\sum_{n =1}^{\infty} \frac{n + \sqrt{n}}{2n^3 -1}$ converges?

My answer: For large n, the given series is smaller than $\sum_{n =1}^{\infty}\frac {2n}{2n^3}$ which is equal to $\sum_{n =1}^{\infty}\frac {1}{n^2}$, but we know that the later is convergent by the p-series test, then by comparison test the given series is convergent.

Is my answer correct?

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    $\begingroup$ It is correct, however you glossed over the exact details of how we know $\frac{n+\sqrt{n}}{2n^3-1}$ is smaller than $\frac{2n}{2n^3}$ for large enough $n$. It would be best to show this more explicitly. $\endgroup$ – JMoravitz Jan 15 at 5:39
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    $\begingroup$ It is correct. In fact the inequality holds for all $n$. Just verify it carefully to make the argument complete. $\endgroup$ – Kabo Murphy Jan 15 at 5:39
  • $\begingroup$ @KaviRamaMurthy I think it does not hold for $n=1$ $\endgroup$ – Secretly Jan 15 at 5:42
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    $\begingroup$ @hopefully Right. It is true for all $n>1$. $\endgroup$ – Kabo Murphy Jan 15 at 5:45
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    $\begingroup$ Also, because the Comparison Test applies only to series with nonnegative terms, you should mention that $\frac{n + \sqrt{n}}{2n^3 -1}>0$. $\endgroup$ – Taladris Jan 15 at 5:49
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Almost correct.

You have to make the denominator smaller, not larger. So yse $n^3$ and everything is fine.

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  • $\begingroup$ could not understand what do you mean? $\endgroup$ – Secretly Jan 15 at 5:47
  • $\begingroup$ I think for large n increasing the denominator by 1 would not matter, because I have increased the numerator also. $\endgroup$ – Secretly Jan 15 at 5:49
  • $\begingroup$ yeah I got it ....great! $\endgroup$ – Secretly Jan 15 at 6:05

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