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Alternative telescoping patterns

Authors
  • avatar
    Name
    Vu Hung
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Problem Statement

  • Evaluate ∑n=04(−1)n+1(n+5)\sum_{n=0}^{4}(-1)^{n+1}(n+5).
  • Show 1n(n+1)=1n−1n+1\frac{1}{n(n+1)}=\frac1n-\frac1{n+1}.
  • Hence find
SN=∑n=1N(1n−1n+1)S_N=\sum_{n=1}^N\left(\frac1n-\frac1{n+1}\right)

and lim⁡N→∞SN\lim_{N\to\infty}S_N.

Hints

Part (iii) is a direct telescoping sum after part (ii).

Solutions

∑n=04(−1)n+1(n+5)=−7.\sum_{n=0}^{4}(-1)^{n+1}(n+5)=-7.

Also

1n(n+1)=(n+1)−nn(n+1)=1n−1n+1.\frac1{n(n+1)}=\frac{(n+1)-n}{n(n+1)}=\frac1n-\frac1{n+1}.

So

SN=1−1N+1,lim⁡N→∞SN=1.S_N=1-\frac1{N+1},\qquad \lim_{N\to\infty}S_N=1.

Further Readings

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