I have to calculate the series \sum_{k=2}^\infty(\frac{1}{k}-\frac{1}{k+2}) Using the definition: L=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=0}^n a^k Obviously \lim_{n\to\infty}(\frac{1}{n}-\frac{1}{n+2})=0,

Mary Reyes 2021-12-22 Answered
I have to calculate the series
k=2(1k1k+2)
Using the definition:
L=limnSn=limnk=0nak
Obviously limn(1n1n+2)=0, but I don't think that this is the right way to calculate the value of the series.
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jean2098
Answered 2021-12-23 Author has 38 answers
Let us use 1k=01tk1dt. Then
S=k=2(1k1k+2)=01k=2[tk1tk+2]dt
=01tt31tdt
=01(t+t2)dt
=12+13
=56
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Foreckije
Answered 2021-12-24 Author has 32 answers
As suggested from the comment by JMoravitz, you should write down the series explicitly and see if you can see something interesting:
2(1k1k+2)=(1214)+(1315)+(1416)+(1517)+(1618)+
From here, you can see which terms are cancelling with one another and which one is left.... Essentially you should see that all the terms will cancel with each other except the 12 and 13.
Therefore, the answer is 56.
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user_27qwe
Answered 2021-12-30 Author has 208 answers

Another, Different way that uses limit like you wanted to is to find the partial sum! You first compute, S2=1214, S3=S2+(1315) and so on... What you will find (and you can prove this fact by induction) is that
Sn=5n2+3n86(n+1)(n+2)
Now,
limn5n2+3n86n2+18n+12=56

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