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Convergence of three series and ∑sin⁡n\displaystyle \sum\sin n

Math I: Calculus · Mock exam, Question 1(b)–(c) · 14 points · Lecture 10

Problem

  • (b) Determine whether ∑n=1∞sin⁡n\displaystyle \sum_{n=1}^{\infty}\sin n converges. Justify your conclusion using a necessary condition for convergence of a series. [4 points]

  • (c) Let cn=1/n.c_n=1/\sqrt{n}\text{.} Determine which of the following series converge and find any accessible sum. [10 points]

    S1=∑n=1∞cn3,S2=∑n=1∞cnn1/3,S3=∑n=1∞n1/2cn(−2)n.\begin{aligned} S_1&=\sum_{n=1}^{\infty}\frac{c_n}{3},\\ S_2&=\sum_{n=1}^{\infty}\frac{c_n}{n^{1/3}},\\ S_3&=\sum_{n=1}^{\infty}\frac{n^{1/2}c_n}{(-2)^n}. \end{aligned}

What you need

  • Infinite series: if ∑bn\displaystyle \sum b_n converges, then bn→0b_n\to0 (the term test).
  • Trigonometric functions: sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1 and the addition formula for sin⁡(x+y).\sin(x+y)\text{.}
  • pp-series: ∑n−p\displaystyle \sum n^{-p} converges exactly when p>1.p>1\text{.}
  • Geometric series: for ∣r∣<1,|r|<1\text{,} ∑n=1∞rn=r/(1−r).\displaystyle \sum_{n=1}^{\infty}r^n=r/(1-r)\text{.}

Plan

In (b) the term test can only prove divergence, so the work is to show that sin⁡n\sin n does not tend to 0.0\text{.} Argue by contradiction with the addition formula. In (c), substitute cnc_n first: each series then becomes a standard one you can recognise.

Step-by-step solution

Try each step yourself before you open it.

(b) Step 1 · State the necessary condition (1 point)

If ∑bn\displaystyle \sum b_n converges, then bn→0.b_n\to0\text{.} So if sin⁡n↛0,\sin n\not\to0\text{,} the series diverges.

Concept: Infinite series

(b) Step 2 · Show that sin n does not tend to 0 (2 points)

Suppose, for contradiction, that sin⁡n→0.\sin n\to0\text{.} Then cos⁡2n=1−sin⁡2n→1,\cos^2n=1-\sin^2n\to1\text{,} so ∣cos⁡n∣→1.|\cos n|\to1\text{.} The addition formula gives

sin⁡(n+1)−sin⁡ncos⁡1=cos⁡nsin⁡1.\sin(n+1)-\sin n\cos1=\cos n\sin1.

The left-hand side tends to 0−0⋅cos⁡1=0,0-0\cdot\cos1=0\text{,} because sin⁡(n+1)→0\sin(n+1)\to0 as well. The absolute value of the right-hand side tends to ∣sin⁡1∣>0.|\sin1|>0\text{.} This is a contradiction, so sin⁡n↛0.\sin n\not\to0\text{.}

(b) Step 3 · Conclude (1 point)

By the term test, ∑n=1∞sin⁡n\displaystyle \sum_{n=1}^{\infty}\sin n diverges.

(c) Step 1 · Rewrite S₁ and S₂ (2 points)

With cn=n−1/2,c_n=n^{-1/2}\text{,}

S1=13∑n=1∞n−1/2,S2=∑n=1∞n−5/6,S_1=\frac{1}{3}\sum_{n=1}^{\infty}n^{-1/2},\qquad S_2=\sum_{n=1}^{\infty}n^{-5/6},

since n−1/2n−1/3=n−5/6.n^{-1/2}n^{-1/3}=n^{-5/6}\text{.}

(c) Step 2 · Apply the p-series test (3 points)

S1S_1 is a constant multiple of the pp-series with p=1/2≤1,p=1/2\leq1\text{,} which diverges. A nonzero constant factor does not change divergence, so S1S_1 diverges.

S2S_2 is the pp-series with p=5/6≤1,p=5/6\leq1\text{,} so it diverges too.

Concept: pp-series

(c) Step 3 · Recognise S₃ as geometric (2 points)

n1/2cn=1,n^{1/2}c_n=1\text{,} so

S3=∑n=1∞1(−2)n=∑n=1∞(−12)n.S_3=\sum_{n=1}^{\infty}\frac{1}{(-2)^n} =\sum_{n=1}^{\infty}\left(-\frac{1}{2}\right)^n.

This is a geometric series with ratio r=−1/2,r=-1/2\text{,} and ∣r∣<1.|r|<1\text{.}

(c) Step 4 · Sum it and classify (3 points)
S3=−1/21−(−1/2)=−1/23/2=−13.S_3=\frac{-1/2}{1-(-1/2)}=\frac{-1/2}{3/2}=-\frac{1}{3}.

So S1S_1 and S2S_2 diverge, and only S3S_3 converges, with sum −1/3.-1/3\text{.}

Concept: Geometric series

Where marks are lost

  • Using the term test to prove convergence. bn→0b_n\to0 is necessary, not sufficient: ∑1/n\displaystyle \sum1/\sqrt{n} diverges although 1/n→0.1/\sqrt{n}\to0\text{.}
  • Asserting "sin⁡n\sin n oscillates" without proof. Part (b) asks for a justification of sin⁡n↛0.\sin n\not\to0\text{.}
  • Starting the geometric sum at n=0.n=0\text{.} The series starts at n=1,n=1\text{,} so the first term is −1/2,-1/2\text{,} not 1.1\text{.}

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Symbols: infinity, square root, sum