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. :(

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.