The summation is: $$\sum_{n=1}^\infty \frac{ \sqrt { n + 1 } - \sqrt n }{n^A}$$
I don't know how to even begin. Hints??
Asked
Active
Viewed 96 times
4
-
I tried to fix your $\TeX$ code, did I correctly? – martini May 22 '13 at 07:10
-
Yes. Thanks. typing via phone is PAIN! – Parth Thakkar May 22 '13 at 07:19
3 Answers
4
Hint: A standard beginning is to multiply top and bottom by $\sqrt{n+1}+\sqrt{n}$.
We end up with $$\sum_{n=1}^\infty \frac{1}{n^A(\sqrt{n+1}+\sqrt{n})}.$$ Now note that $\sqrt{n}\lt \sqrt{n+1}\le 2\sqrt{n}$. From this we obtain $$ \frac{1}{3n^{A+1/2}} \lt \frac{1}{n^A(\sqrt{n+1}+\sqrt{n})}\lt \frac{1}{2n^{A+1/2}}.$$ These inequalities should be enough to do the appropriate comparisons with familiar series.
André Nicolas
- 507,029
-
I did the first step and was wondering if somehow inequalities would help (as done in Sandwich theorem), but didn't know how to get further. The inequality helped! Thanks. – Parth Thakkar May 24 '13 at 13:56
-
You are welcome. The first inequality above shows, by Comparison, that we have divergence if $A\ge 1/2$. The second inequality shows there is convergence if $A\lt 1/2$. – André Nicolas May 24 '13 at 14:26
1
HINT:
$$\frac{\sqrt{n+1}-\sqrt{n}}{n^A}=\frac{\sqrt{n+1}-\sqrt{n}}{n^A}\cdot\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\frac1{n^A\left(\sqrt{n+1}+\sqrt{n}\right)}$$
Brian M. Scott
- 616,228
1
In addition to both of neat answers, you could use the following facts to find proper values of $A$ as well:
Let $\lim_{n\to\infty}~n^pu_n=A$. Then:
If $p>1$ and $A$ is finite then $\sum u_n$ converges.
If $p\le1$ and $A\neq0$ or $A=\infty$ then $\sum u_n$ diverges.
-
-
This seems interesting, though never had heard anything before about this. Can you please provide further details, or links to them ?? – Parth Thakkar May 24 '13 at 14:02
-
@ParthThakkar: If you have any access to Schaum's series, see Advanced Calculus. – Mikasa May 25 '13 at 12:03