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Question: Is the series convergent or divergent? $$\sum_{n=0}^{\infty}{\frac{1}{\sqrt{n+1}}}$$

I can use any test but wolfram alpha says that it is divergent by comparison test.

How do I apply comparison test?

I can compare it to: $$\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ \sqrt { n } } }$$ but the second series is greater than the series in the question and the second series is divegent. :(

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  • $\begingroup$ They are the same series, written in different forms. $\endgroup$ – Spook Apr 15 '13 at 16:15
  • $\begingroup$ Use $\sum \frac{1}{2 \sqrt{n}}$ then... $\endgroup$ – vonbrand Apr 15 '13 at 17:27
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Rewrite the first series with the substitution $k=n+1$, yielding $$\sum_{k=1}^\infty\frac1{\sqrt k}.$$

The series $$\sum_{n=0}^\infty\frac1{\sqrt n}$$ makes no sense, since $\frac1{\sqrt n}$ is undefined for $n=0$.


Alternately, you could use the comparison test as follows. For $n\ge1,$ $$\frac1{\sqrt{n+1}}\ge\frac1{\sqrt{2n}}=\frac1{\sqrt2}\frac1{\sqrt n},$$ so that $$\begin{align}\sum_{n=0}^\infty\frac1{\sqrt{n+1}} &= 1+\sum_{n=1}^\infty\frac1{\sqrt{n+1}}\\ &\ge 1+\frac1{\sqrt2}\sum_{n=1}^\infty\frac1{\sqrt n},\end{align}$$ so since $\sum_{n=1}^\infty\frac1{\sqrt n}$ diverges, so does the series we're considering.

Reindexing is certainly the neatest trick, here, though.

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  • $\begingroup$ Wow. Unbelievable. I feel dumb. My textbook didn't teach me this method. Thank you so much! $\endgroup$ – user72708 Apr 15 '13 at 16:16
  • $\begingroup$ No problem. See my updated answer for another way to go about it, using the comparison test (this may be how W|A proceeded). $\endgroup$ – Cameron Buie Apr 15 '13 at 16:48
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For large enough $n$, $\sqrt{n + 1} < n$, so that $\dfrac{1}{\sqrt{n + 1}} > \dfrac{1}{n}$, and as the harmonic series diverges, so does yours.

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