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

Mary Reyes

Answered question

2021-12-22

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.

Answer & Explanation

jean2098

jean2098

Beginner2021-12-23Added 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
Foreckije

Foreckije

Beginner2021-12-24Added 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.
user_27qwe

user_27qwe

Skilled2021-12-30Added 375 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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