# calculus

i have to determine whether the series is convergent, and if, find the sum

the series is from k=1 to infinity of
2/((k+1)(k+3))

I got 5/6 as my answer and didn't know if it was right...

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1. There are many tests for convergence. In this case you can compare the series to the known convergent series 1/k^2 from k = 1 to infinity. Since
2/((k+1)(k+3)) is always less than 2/k^2 the series must converge.

The summation can be computed by writing:

2/((k+1)(k+3)) =

1/(k+1) - 1/(k+3)

So, you would expect the summation to be given by the sum of 1/(k+1) from k = 1 to k = 2 (the two terms that are not conceled by 1/(k+3)), which is the answer you obtained.

Since the two separate summations do not converge you must prove that this is indeed the case. What you do is you say that the summation from 1 to infinity of 2/((k+1)(k+3)) equals the limit of N goes to infinity of the summation from 1 to N. For finite N you then evaluate the summation of

1/(k+1) - 1/(k+3)

from k = 1 to N.

You then get four terms: the k = 1 and k = 2 terms of 1/(k+1) which yields 5/6 and the k = N-1 and k = N terms of
- 1/(k+3). So, you get:

5/6 - 1/(N+2) - 1/(N+3)

Finally, you take the limit
N --> infinity which yields the answer of 5/6

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2. The terms can be rewritten
1/(k+1) -1/(k+3)
I used the technique of partial sums. You can verify that the two expressions are equivalent.
Summing from x=1 to infinity, all terms will cancel in pairs except 1/2 + 1/3, for k=1 and k=2. The sum of those terms is 5/6, so I agree.

Did you use a different method?

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3. kinda helpful

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