Question

Given are two infinite series
P: $\displaystyle \sum \frac{n^2 + 1}{n^2}$
Q: $\displaystyle \sum \left(1 + \frac{1}{n}\right)^{-n}$
The correct choice is:

• A. P is convergent series; Q is divergent series
• B. P is divergent series; Q is convergent series
• C. Both P and Q are convergent series
• D. Both P and Q are divergent series

Hint:

Always check whether the term of the series tends to zero as $n \to \infty$. If not, the series must diverge.

Answer 1

The Correct Option is B

Step 1: Analyze series P.
\[ \frac{n^2 + 1}{n^2} = 1 + \frac{1}{n^2} \] So the series becomes: \[ \sum \left(1 + \frac{1}{n^2}\right) = \sum 1 + \sum \frac{1}{n^2} \] The term $\sum 1$ diverges (infinite sum of ones). Therefore, P is a divergent series.
Step 2: Analyze series Q.
\[ \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{-n} = \frac{1}{e} \] So each term tends to $\dfrac{1}{e}$, not $0$. Wait carefully — but note: For convergence of $\sum a_n$, we require $a_n \to 0$. Here, $a_n \to \dfrac{1}{e} \neq 0$, so the series diverges. Correction: Actually, Q is divergent. So final: (D) Both P and Q are divergent. Final Answer: \[ \boxed{\text{(D) Both P and Q are divergent series}} \]