Question

Consider the two series, \(S_A\) and \(S_B\), where
\[ S_A=\sum_{n=1}^{\infty}\frac{n^2}{2^n} \]
\[ S_B=1+\frac{1}{2}+\frac{1}{8}+\frac{1}{16}+\frac{1}{64}+\frac{1}{128}+\frac{1}{512}+\cdots \]
Which of the following statements is correct for the two given series?

• A. Both \(S_A\) and \(S_B\) converge.
• B. Neither \(S_A\) nor \(S_B\) converges.
• C. \(S_A\) converges but \(S_B\) does not converge.
• D. \(S_B\) converges but \(S_A\) does not converge.

Hint:

Check S_A with the ratio test, then split S_B into two interleaved geometric series with common ratio 1/8 each.

Answer 1

The Correct Option is A

Step 1: Test S_A for convergence using the ratio test.
For \(S_A=\sum_{n=1}^{\infty}\frac{n^2}{2^n}\), apply the ratio test with \(a_n=\frac{n^2}{2^n}\):
\[ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\frac{(n+1)^2}{2^{n+1}}\times\frac{2^n}{n^2}=\lim_{n\to\infty}\frac{1}{2}\left(\frac{n+1}{n}\right)^2=\frac{1}{2} \]
Since the limit \(\frac{1}{2}\) is less than \(1\), the series \(S_A\) converges.

Step 2: Find the exact value of S_A as a check.
Using the standard identity \(\sum_{n=1}^{\infty}n^2x^n=\frac{x(1+x)}{(1-x)^3}\) at \(x=\frac{1}{2}\):
\[ S_A=\frac{\frac{1}{2}\left(1+\frac{1}{2}\right)}{\left(1-\frac{1}{2}\right)^3}=\frac{\frac{1}{2}\times\frac{3}{2}}{\frac{1}{8}}=\frac{\frac{3}{4}}{\frac{1}{8}}=6 \]
So \(S_A\) converges to a finite value, \(6\).

Step 3: Spot the pattern in S_B.
Write out the terms of \(S_B\) and look at the denominators:
\[ 1,\ \frac{1}{2},\ \frac{1}{8},\ \frac{1}{16},\ \frac{1}{64},\ \frac{1}{128},\ \frac{1}{512},\ldots \]
Splitting into odd-position terms (1st, 3rd, 5th, ...) and even-position terms (2nd, 4th, 6th, ...): odd-position terms \(1,\ \frac{1}{8},\ \frac{1}{64},\ldots\) each equal to \(\frac{1}{8}\) times the previous one, and even-position terms \(\frac{1}{2},\ \frac{1}{16},\ \frac{1}{128},\ldots\) also each equal to \(\frac{1}{8}\) times the previous one.

Step 4: Sum the two interleaved geometric series.
The odd-position terms form a geometric series with first term \(1\) and common ratio \(\frac{1}{8}\):
\[ 1+\frac{1}{8}+\frac{1}{64}+\cdots=\frac{1}{1-\frac{1}{8}}=\frac{8}{7} \]
The even-position terms form a geometric series with first term \(\frac{1}{2}\) and common ratio \(\frac{1}{8}\):
\[ \frac{1}{2}+\frac{1}{16}+\frac{1}{128}+\cdots=\frac{\frac{1}{2}}{1-\frac{1}{8}}=\frac{1}{2}\times\frac{8}{7}=\frac{4}{7} \]

Step 5: Add both parts.
\[ S_B=\frac{8}{7}+\frac{4}{7}=\frac{12}{7} \]
Since \(\frac{12}{7}\) is a finite number, \(S_B\) also converges.

Step 6: Analyze the options.

(A) Both S_A and S_B converge: True, as shown above.

(B) Neither S_A nor S_B converges: False, both converge.

(C) S_A converges but S_B does not converge: False, S_B also converges.

(D) S_B converges but S_A does not converge: False, S_A also converges.

Step 7: Final conclusion.
Therefore, the correct statement is \[ \boxed{\text{Both }S_A\text{ and }S_B\text{ converge.}} \]