Evaluate Telescoping Sums:a/b for Integer k>0

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In summary, we can evaluate the telescoping sums for (a) and (b) to be $\frac{1}{4}$ and $\frac{1}{k}(\frac{1}{1}+\ldots+\frac{1}{k})$, respectively.
  • #1
alexmahone
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Evaluate using telescoping sums:

(a) $\sum_1^\infty\frac{(-1)^{n-1}}{n(n+2)}$

(b) $\sum_1^\infty\frac{1}{n(n+k)}$, $k$ integer $>0$

My attempt:

(a)$\frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)$

Adding the terms for $n$ even, we get

$-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\ldots\right)=-\frac{1}{4}$

Adding the terms for $n$ odd, we get

$\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\ldots\right)=\frac{1}{2}$

So, the total is $-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}$

(b) $\sum_1^\infty\frac{1}{n(n+k)}=\sum_1^\infty\frac{1}{k}\left(\frac{1}{n}-\frac{1}{n+k}\right)$

$=\frac{1}{k}\left(\frac{1}{1}-\frac{1}{1+k}+\frac{1}{2}-\frac{1}{2+k}+\frac{1}{3}-\frac{1}{3+k}+\ldots\right)$

How do I proceed?
 
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  • #2
Hello,

****************** latex. I hate this version...Forget it for the first one... But please post your thing at once, because you're constantly editing and it's really ticking me off to try writing something that yourself are currently writing !

As for the second one, reason as if k=2, how would you do ? Then it'd be the same except that it's a little more, but in the end they'll all cancel each other out.
 
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  • #3
Moo said:
As for the second one, reason as if k=2, how would you do ? Then it'd be the same except that it's a little more, but in the end they'll all cancel each other out.

I think the answer is $\frac{1}{k}(\frac{1}{1}+\ldots+\frac{1}{k})$. Can that be simplified further?
 
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  • #4
Yes, that's the answer, and no further simplification follows, since the harmonic sum doesn't have much nice properties.
 

FAQ: Evaluate Telescoping Sums:a/b for Integer k>0

What is a telescoping sum?

A telescoping sum is a series of numbers where most of the terms cancel each other out, leaving only a few terms that cannot be simplified.

How do you evaluate a telescoping sum?

To evaluate a telescoping sum, you must first identify the pattern of terms that cancel each other out. Then, you can simplify the remaining terms and find the sum.

What does "a/b" represent in the telescoping sum formula?

"a/b" represents the general term of the telescoping sum. It is usually a fraction or a polynomial with variable terms.

What is the significance of "k" being an integer greater than 0 in the telescoping sum?

"k" represents the number of terms in the telescoping sum. It must be an integer greater than 0 in order for the telescoping sum to have a finite value.

Can telescoping sums be used in real-world applications?

Yes, telescoping sums can be used in real-world applications such as calculating the cost of a project with recurring expenses or finding the total distance traveled in a journey with changing speeds.

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