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: Apply the root test to S_A.
For $S_A=\sum \frac{n^2}{2^n}$, look at $\left(\dfrac{n^2}{2^n}\right)^{1/n}=\dfrac{n^{2/n}}{2}$. As $n\to\infty$, $n^{2/n}\to1$, so the limit is $\dfrac{1}{2}$, which is below $1$. By the root test, $S_A$ converges.

Step 2: Group the terms of S_B in pairs.
Pair up the terms of $S_B$ two at a time:
\[ \left(1+\frac{1}{2}\right)+\left(\frac{1}{8}+\frac{1}{16}\right)+\left(\frac{1}{64}+\frac{1}{128}\right)+\cdots \]
The first pair sums to $\frac{3}{2}$, the second pair sums to $\frac{3}{16}$, the third pair sums to $\frac{3}{128}$.

Step 3: Spot the ratio between pair sums.
\[ \frac{3/16}{3/2}=\frac{1}{8},\qquad \frac{3/128}{3/16}=\frac{1}{8} \]
So the pair sums form a geometric series with first term $\frac{3}{2}$ and ratio $\frac{1}{8}$.

Step 4: Sum this geometric series.
\[ S_B=\frac{3/2}{1-1/8}=\frac{3/2}{7/8}=\frac{3}{2}\times\frac{8}{7}=\frac{12}{7} \]
This is a finite number, so $S_B$ converges too.

Step 5: Conclude.
Since both series settle to finite values, $6$ for $S_A$ and $\frac{12}{7}$ for $S_B$, the correct statement is that both series converge.
\[ \boxed{\text{Both }S_A\text{ and }S_B\text{ converge.}} \]