Question

Show that the series

n=1an\sum_{n=1}^{\infty} a_n

can be written in the telescoping form

n=1[(cSn1)(cSn)]\sum_{n=1}^{\infty}\left[\left(c-S_{n-1}\right)-\left(c-S_n\right)\right]

where S0=0S_0=0 and SnS_n is the nth partial sum.

Solution

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Answered 3 years ago
Answered 3 years ago
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Since the parcial sum is defined as Sn=k=1nakS_n=\displaystyle{\sum_{k=1}^{n}} a_k and Sn1=k=1n1akS_{n-1}=\displaystyle{\sum_{k=1}^{n-1}} a_k it means that

SnSn1=k=1nakk=1n1ak=anS_n - S_{n-1}=\sum_{k=1}^{n} a_k - \sum_{k=1}^{n-1} a_k=a_n

Hence,

an=SnSn1=Snc+cSn1=(cSn)(cSn1)a_n =S_n - S_{n-1} = S_n -c+c- S_{n-1}=\left(c- S_n \right) - \left( c-S_{n-1}\right)

Now let's rewrite the initial series

n=1an=n=1[(cSn)(cSn1)]\boldsymbol{\sum_{n=1}^{\infty} a_n } = \boldsymbol{\sum_{n=1}^{\infty} \left[\left(c - S_n \right) - \left(c - S_{n-1}\right) \right] }

And this is the telescoping form of the series.

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