← 2022 Paper 2

UPSC 2022 Maths Optional Paper 2 Q4b — Step-by-Step Solution

15 marks · Section A

Series of real terms: convergence, standard tests · Real Analysis · asked 2× in 13 yrs · Read the full method →

Question

Test for convergence or divergence of $x+\dfrac{2^2 x^2}{2!}+\dfrac{3^3 x^3}{3!}+\cdots+\dfrac{n^n x^n}{n!}+\cdots$ ($x>0$).

Technique

Ratio test; uses $(1+1/n)^n\to e$; Stirling for the borderline case.

Solution

Setup. $a_n=\dfrac{n^n x^n}{n!}$.

Step 1 — Apply ratio test

Step 2 — Take limit as $n\to\infty$

(Using $\lim(1+1/n)^n=e$.)

Step 3 — Apply ratio test criterion

• $xe<1$, i.e., $x<1/e$: series converges.
• $xe>1$, i.e., $x>1/e$: series diverges.
• $x=1/e$: inconclusive (need another test).

Step 4 — Handle borderline $x=1/e$

At $x=1/e$: $a_n=\dfrac{n^n}{n!\cdot e^n}$.

Use Stirling’s approximation: $n!\sim\sqrt{2\pi n}\cdot n^n e^{-n}$.

So $a_n\sim\dfrac{n^n}{\sqrt{2\pi n}\cdot n^n e^{-n}\cdot e^n}=\dfrac{1}{\sqrt{2\pi n}}$.

Series $\sum 1/\sqrt n$ diverges. So at $x=1/e$, series diverges.

Conclusion

Answer