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Show that each of the series converges on their respective domains.

a) $$\sum_{n=1}^\infty \frac{1}{(1 + nx)^2}, x \in (0,\infty)$$

b)$$\sum_{n=1}^\infty e^{-nx}, x \in(0,\infty)$$

For the first one, I have said that $$\sum_{n=1}^\infty \frac{1}{(1 + nx)^2} < \sum_{n=1}^\infty \frac{1}{ n^2}$$ for all $x \in(0,\infty)$ and since $\sum_{n=1}^\infty \frac{1}{ n^2}$ converges then by the comparison test $\sum_{n=1}^\infty \frac{1}{(1 + nx)^2}$ converges for all $x>0$

For the second one, I have said that $$\sum_{n=1}^\infty e^{-nx} < \sum_{n=1}^\infty e^{-n} = \sum_{n=1}^\infty \left(\frac{1}{e}\right)^n$$ and since $\sum_{n=1}^\infty \left(\frac{1}{e}\right)^n$ converges because its a geometric series, then by the comparison test ,the series $\sum_{n=1}^\infty e^{-nx}$ converges for all $x>0$

Please tell me if this is correct or what is the correct way for these problems.

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2  
For the first, say it is $\lt \frac{1}{x^2}\cdot\frac{1}{n^2}$. For the second, don't throw away the $x$, you have a geometric series. –  André Nicolas Apr 25 at 22:41
    
@André Nicolas ok thanks! –  user143612 Apr 25 at 22:51
1  
I dont know how to comment on a user's answer, but thanks to Alex Zorn and Nicholas Stull –  user143612 Apr 25 at 22:57
    
@AndréNicolas can I say $\sum_{n=1}^\infty \frac{1}{n^2}$ converges so $\frac{1}{x^2}\sum_{n=1}^\infty \frac{1}{n^2}$ also converges? –  user143612 Apr 25 at 23:07
    
Yes, $x$ us fixed, so $1/x^2$ is a constant. –  André Nicolas Apr 25 at 23:22

2 Answers 2

It's not true that:

$$\frac{1}{(1 + nx)^2} < \frac{1}{n^2}$$

You'll want to take Andre's suggestion and do:

$$\frac{1}{(1 + nx)^2} < \frac{1}{(nx)^2}$$

For the second one, it's also not true that

$$e^{-nx} < e^{-n}$$

For the second one, you don't even need to do a comparison test: The series is already a geometric series with common ratio $e^{-x}$.

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Hint on the first:

As Andre suggested, use instead $$\frac{1}{(1+nx)^2} \leq \frac{1}{(nx)^2}$$ and now you can write the following: $$\sum_{n=1}^\infty \frac{1}{(1+nx)^2} < \sum_{n=1}^\infty \frac{1}{(nx)^2} = \frac{1}{x^2}\sum_{n=1}^\infty \frac{1}{n^2}$$ Note you need $x > 0$ (or at least $x\neq 0$) for the term $\frac{1}{x^2}$ to be finite, and from there, what do you know about the series written in that last step?

Hint on the second one:

$$\sum_{n=1}^\infty e^{-nx} = \sum_{n=1}^\infty \left(e^{-x}\right)^n$$

Now, this is a geometric series. When do geometric series converge? (and what would this mean in terms of $x$?)

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