Proving Convergence of Infinite Series with Changing Signs

ait.abd
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Homework Statement


Show the convergence of the series
\sum_{n=1}^{inf}(\frac{1}{n}-\frac{1}{n+x})
of real-valued functions on R - \{-1, -2, -3, ...\}.

Homework Equations


The Attempt at a Solution


I first thought of solving this using telescoping series concept but it didn't work out. Also, I tried to prove it by saying that since denominator goes n^2 it should converge but what I doubt is that the elements of the series change sign and the modulus of its terms will not be lower than 1/n^2.
 
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Well, you could try to turn this in a telescoping series.

Pick m a natural number such that x<m. Then

\frac{1}{n}-\frac{1}{n+x}\leq \frac{1}{n}-\frac{1}{n+m}...
 
micromass said:
Well, you could try to turn this in a telescoping series.

Pick m a natural number such that x<m. Then

\frac{1}{n}-\frac{1}{n+x}\leq \frac{1}{n}-\frac{1}{n+m}...

Thanks micromass!
 
I have a slight problem with the solution micromass hope you can clarify. I was looking at the statement of comparison tests that states that a_n, b_n &gt; 0 for comparison test to be valid whereas in the solution above we can have negative individual terms as well? Is it so or I am looking at it wrongly?
e.g.
n=1, x=-0.1 \implies 1-10/9 &lt; 0
 
Given the series:

1/1 + 1/(1+2) + 1/(1+2+3) + 1/(1+2+3+4) + ...

How would u find the sum of the series?
Any help pleaseee!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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