I have to calculate the series∑k=2∞(1k−1k+2)Using the definition:L=limn→∞Sn=limn→∞∑k=0nakObviously limn→∞(1n−1n+2)=0, but I don't think that this is the right way to calculate the value of the series.

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I have to calculate the series∑k=2∞(1k−1k+2)Using the definition:L=limn→∞Sn=limn→∞∑k=0nakObviously limn→∞(1n−1n+2)=0, but I don&#039;t think that this is the right way to calculate the value of the series.

jean2098 · Accepted Answer

Let us use 1k=∫01tk−1dt. Then  S=∑k=2∞(1k−1k+2)=∫01∑k=2∞[tk−1−tk+2]dt  =∫01t−t31−tdt  =∫01(t+t2)dt  =12+13  =56

Foreckije · Accepted Answer

As suggested from the comment by JMoravitz, you should write down the series explicitly and see if you can see something interesting:  ∑2∞(1k−1k+2)=(12−14)+(13−15)+(14−16)+(15−17)+(16−18)+…  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 · Accepted Answer

Another, Different way that uses limit like you wanted to is to find the partial sum! You first compute, S2=12−14, S3=S2+(13−15) and so on... What you will find (and you can prove this fact by induction) is that Sn=5n2+3n−86(n+1)(n+2) Now, limn→∞5n2+3n−86n2+18n+12=56

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,

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