GATE 2022 EC Q.45: solution and answer

Official answer: B, DSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

1. Let $a_n=\frac{n^d}{c^n}$. Every option has $c>0$, so every term is positive. Apply the ratio test: $$\frac{a_{n+1}}{a_n}={\left(\frac{n+1}{n}\right)}^d\cdot \frac{1}{c}⟶\frac{1}{c}(n\to \infty ),$$ for any value of $d$, because ${\left(\frac{n+1}{n}\right)}^d\to 1$.

2. So the series converges if $c>1$ (the limit is less than $1$) and diverges if $0<c<1$ (the limit is greater than $1$), for every $d$. A power $n^d$ can never outgrow an exponential $c^n$, and can never hold one back.

3. When $c=1$ the ratio test gives a limit of exactly $1$, so it tells us nothing. The series becomes $\sum n^d=\sum \frac{1}{n^{-d}}$, a p-series with $p=-d$. It converges if and only if $p>1$, that is, $d<-1$.

4. (A) $c=1,d=-1$: $\sum \frac{1}{n}$ is the harmonic series ($p=1$), which diverges.

5. (B) $c=2,d=1$: $\sum \frac{n}{2^n}$ has ratio limit $\frac{1}{2}<1$, so it converges. (Its sum is $2$.)

6. (C) $c=0.5,d=-10$: $a_n=\frac{n^{-10}}{(1/2{)}^n}=\frac{2^n}{n^{10}}\to \infty$, and the ratio limit is $2>1$. The terms do not even go to zero (the $n$-th term test), so the series diverges.

7. (D) $c=1,d=-2$: $\sum \frac{1}{n^2}$ is a p-series with $p=2>1$, so it converges.

8. The series converges for (B) and (D).