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Telescoping sums and inductive proof

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    Name
    Vu Hung
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Problem Statement

Consider the sequence

ak=2k(k+1)(k+2)(k≥1).a_k=\frac{2}{k(k+1)(k+2)}\qquad (k\ge 1).

Define the partial sums

Sn=∑k=1nak.S_n=\sum_{k=1}^n a_k.
  • Show, using partial fractions, that
2k(k+1)(k+2)=1k(k+1)−1(k+1)(k+2).\frac{2}{k(k+1)(k+2)} =\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}.
  • Demonstrate that the sequence of partial sums telescopes. Hence find a simplified expression for SnS_n in terms of nn.
  • Prove the expression for SnS_n found in part (ii) by mathematical induction for all n≥1n\ge 1.
  • Determine the limiting sum of the series as n→∞n\to\infty.

Hints

  • For (i): Put the two fractions over the common denominator k(k+1)(k+2)k(k+1)(k+2).
  • For (ii): Write out the first few terms of SnS_n and look for cancellation.
  • For (iii): Use Sm+1=Sm+am+1S_{m+1}=S_m+a_{m+1} in the induction step.
  • For (iv): Take the limit of the closed form for SnS_n.

Solutions

(i)

1k(k+1)−1(k+1)(k+2)=k+2−kk(k+1)(k+2)=2k(k+1)(k+2).\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)} =\frac{k+2-k}{k(k+1)(k+2)} =\frac{2}{k(k+1)(k+2)}.

(ii) Therefore

ak=1k(k+1)−1(k+1)(k+2).a_k=\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}.

So

Sn=(11⋅2−12⋅3)+(12⋅3−13⋅4)+⋯+(1n(n+1)−1(n+1)(n+2))=12−1(n+1)(n+2).\begin{aligned} S_n &=\left(\frac1{1\cdot2}-\frac1{2\cdot3}\right) +\left(\frac1{2\cdot3}-\frac1{3\cdot4}\right) +\cdots +\left(\frac1{n(n+1)}-\frac1{(n+1)(n+2)}\right)\\ &=\frac12-\frac{1}{(n+1)(n+2)}. \end{aligned}

(iii) We prove

Sn=12−1(n+1)(n+2).S_n=\frac12-\frac{1}{(n+1)(n+2)}.

For n=1n=1,

S1=21⋅2⋅3=13S_1=\frac{2}{1\cdot2\cdot3}=\frac13

and

12−12⋅3=12−16=13.\frac12-\frac1{2\cdot3}=\frac12-\frac16=\frac13.

Now assume the formula holds for n=mn=m. Then

Sm+1=Sm+2(m+1)(m+2)(m+3)=12−1(m+1)(m+2)+(1(m+1)(m+2)−1(m+2)(m+3))=12−1(m+2)(m+3).\begin{aligned} S_{m+1} &=S_m+\frac{2}{(m+1)(m+2)(m+3)}\\ &=\frac12-\frac1{(m+1)(m+2)} +\left(\frac1{(m+1)(m+2)}-\frac1{(m+2)(m+3)}\right)\\ &=\frac12-\frac1{(m+2)(m+3)}. \end{aligned}

This is the required formula with n=m+1n=m+1, so the result holds for all n≥1n\ge1.

(iv)

lim⁡n→∞Sn=lim⁡n→∞(12−1(n+1)(n+2))=12.\lim_{n\to\infty}S_n =\lim_{n\to\infty}\left(\frac12-\frac{1}{(n+1)(n+2)}\right) =\frac12.

Takeaways

  • Telescoping sums work by rewriting terms so that most neighbouring parts cancel.
  • Induction is a useful way to verify a closed form found by pattern spotting.
  • A finite partial-sum formula can make the limiting sum immediate.

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