Series - The Basics — Question 10

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Question 10

For a real parameter rr, consider ∑n=1∞r2n\displaystyle\sum_{n=1}^{\infty}r^{2n}.

  1. Identify the first term and common ratio as functions of rr.

  2. Determine all rr for which the series converges and find its sum.

  3. Analyze the boundary values r=±1r=\pm1 and the region |r|>1|r|>1 using the nth-term test.

Original worksheet page 1: question and worked solution for 4-3-010
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Question 10 – Solution

Step 1: Identify the geometric data.

Expanding the first terms gives r2+r4+r6+⋯,r^2+r^4+r^6+\cdots, Thus the first term is a=r2a=r^2 and the common ratio is q=r2q=r^2. A geometric series converges precisely when |q|<1|q|<1, so |r2|<1⇔|r|<1.|r^2|<1\iff |r|<1.

Step 2: Find the sum on the convergence interval.

For |r|<1|r|<1, the geometric formula gives S(r)=r21−r2.S(r)=\frac{r^2}{1-r^2}. The finite partial sum provides a direct verification: sN=r2(1−r2N)1−r2(r2≠1),s_N=\frac{r^2(1-r^{2N})}{1-r^2}\qquad(r^2\ne1), When |r|<1|r|<1, r2N→0r^{2N}\to0, so sN→r2/(1−r2)s_N\to r^2/(1-r^2).

Step 3: Check the endpoints and exterior region.

At both r=1r=1 and r=−1r=-1, every term r2nr^{2n} equals 11, so sN=N→∞s_N=N\to\infty. If |r|>1|r|>1, then r2nr^{2n} grows rather than approaches zero; the nth-term test proves divergence. Thus the convergence set is exactly (−1,1)(-1,1).

Original worksheet page 2: question and worked solution for 4-3-010

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