Question

Consider the following series: \[ \sum_{n=1}^{\infty} \frac{n^{d}}{c^{n}} \] For which of the following combinations of $c, d$ values does this series converge?

• A. $c = 1,\ d = -1$
• B. $c = 2,\ d = 1$
• C. $c = 0.5,\ d = -10$
• D. $c = 1,\ d = -2$

Hint:

For series of the form $\sum n^{d}/c^{n}$, exponential decay dominates polynomial growth whenever $c>1$. The only exception is when $c=1$, where the series reduces to $\sum n^{d}$, a p-series that converges only if $d<-1$.

Answer 1

The Correct Option is B, D

The given series is: \[ \sum_{n=1}^{\infty} \frac{n^{d}}{c^{n}}. \] We analyze convergence using the ratio test, which is effective for series involving exponentials.
Step 1: Apply the ratio test. Consider \[ a_n = \frac{n^d}{c^n}. \] Then, \[ \frac{a_{n+1}}{a_n} = \frac{(n+1)^d / c^{n+1}}{n^d / c^n} = \frac{(n+1)^d}{n^d} \cdot \frac{1}{c} = \left(1 + \frac{1}{n}\right)^d \cdot \frac{1}{c}. \] As \(n \to \infty\): \[ \left(1 + \frac{1}{n}\right)^d \to 1. \] Thus, \[ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{c}. \] Step 2: Ratio test condition. The series converges if \[ \frac{1}{c}<1 \quad \Rightarrow \quad c>1. \] Therefore, any value of \(d\) is acceptable, but \(c\) must be strictly greater than 1.
Step 3: Evaluate each option.
(A) \(c = 1\): Diverges (ratio = 1).
(B) \(c = 2>1\): Converges for any \(d\). ✓
(C) \(c = 0.5<1\): Diverges (ratio>1).
(D) \(c = 1\): Usually diverges, but note: If \(d<-1\), \[ \frac{n^d}{1^n} = n^d \] is a p-series with \(p = -d>1\), which converges. Here \(d = -2\), so the series becomes \[ \sum n^{-2}, \] which converges. ✓
Thus, convergent cases are (B) and (D).
Final Answer: (B), (D)