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Find the values of x for which the series ∑K=0∞sinxkconverges.

Short Answer

Expert verified

The series ∑K=0∞sinxkconverges for all real numbers except for x∈2n+1π2|n∈ℕ.

Step by step solution

01

Step 1. Given information.

Given a series ∑K=0∞sinxk.

02

Step 2. Find all values of x for which the series converges.

A geometric series is of the form ∑k=0∞crkfor some constants c and r.

Supposer is a non-zero real number, then ∑k=0∞crk converges to c1-rif and only if r<1.

Here, the series localid="1648832395698" ∑K=0∞sinxkhas r=sinx.

For the series to converge, sinx<1.

Note that -1≤sinx≤1.

So, sinx=1if and only if localid="1648832424318" x=2n+1Ï€2 for all natural number n.

It follows that ∑K=0∞sinxkconverges for all real number except for x=2n+1π2where localid="1648832538629" n∈ℕ.

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