4

Show that $$ \sum\frac{1}{\ln n}(\sqrt{n+1} - \sqrt{n})$$

Converges.

I've tried the telescopic property or even write it as

$$\sum \frac{1}{\ln n (\sqrt{n+1}+\sqrt{n})}$$

But didnt help.

Thanks in advance!

Giiovanna
  • 3,197

4 Answers4

5

It diverges, because for large $n$ you have:

$$\frac{1}{n} \leq \frac{1}{\ln n(\sqrt{n+1}+\sqrt{n})}.$$

agha
  • 10,038
3

It diverges. By comparison test, \begin{align} \frac{\sqrt{n+1} - \sqrt{n}}{\ln{n}} &= \frac{1}{\ln{n}\left(\sqrt{n+1} + \sqrt{n} \right)} \\ &> \frac{1}{n} \end{align} for $n > 100$ (say), and the harmonic series diverges.

Karl Winsor
  • 173
3

Hint(for using telescopic property):

$$\dfrac{1}{\ln(n)}(\sqrt{n+1} - \sqrt{n}) \geq \dfrac{\sqrt{n+1}}{\ln(n+1)} - \dfrac{\sqrt{n}}{\ln(n)}$$

Petite Etincelle
  • 14,767
  • Liu, it is hard to find the limit of the seque ce $\frac{\sqrt{n+1}}{ln(n+1)} - \frac{\sqrt{2}}{ln(2)}$, but if we think it as s real function, we can show that it diverges using L'Hospital. Is that true that if the function diverges, then the serie diverges? – Giiovanna Aug 19 '14 at 23:34
  • @Giiovanna Do you know $\lim_{x\to +\infty}\frac{x^\alpha}{\ln(x)} = +\infty, \forall \alpha >0$ – Petite Etincelle Aug 20 '14 at 07:34
  • I do, but for x real. Does it holds for the sequence (I mean, n natural?) Is there any theorem that relation this two thing? – Giiovanna Aug 20 '14 at 09:48
  • @Giiovanna yes of course, if something converges to a limit, then its subsequence convergs to the same limit – Petite Etincelle Aug 20 '14 at 09:49
  • But we are talking about limit pf functions, we dont have sequences properly – Giiovanna Aug 20 '14 at 09:50
  • Suppose we define f(x) =1 if x is natural and f(x)=x elsewhere. If we take f(n) as sequence, we have it converges, but the limit of f does not exists. – Giiovanna Aug 20 '14 at 09:57
  • @Giiovanna Yeah you are right, but it's not a counterexample of what I said. I mean if something convergs, then..... Your example shows if something doesn't converge, it may have a convergent subsequence. – Petite Etincelle Aug 20 '14 at 10:18
  • @Giiovanna Btw, remark that for a function $f(x)$, $\lim_{x\to a}f(x)=L$ is equivalent to say that for any $x_n \to a$, we have $f(x_n) \to L$ – Petite Etincelle Aug 20 '14 at 10:20
0

Sorry, but it diverges, sinces the general term is equivalent to $\frac{1}{n^{1/2}\ln(n)}$ (Bertrand series).

Tig la Pomme
  • 641