### CSIR JUNE 2011 PART B QUESTION 37 SOLUTION ($\sum\limits_{n=0}^{\infty}2^{-n}z^{2n}$)

The power series $\sum\limits_{n=0}^{\infty}2^{-n}z^{2n}$ converges if
1)$|z| \le 2$,
2)$|z|<2$,
3)$|z| \le \sqrt 2$,
4) $|z|<\sqrt 2$.
Solution: Let $f(z) = \sum\limits_{n=0}^{\infty}2^{-n}z^{2n}$ be the analytic function defined on whenever the given series converges. We know that a power series $\sum\limits_{n=0}^{\infty} a_n (z-z_0)^n$ has the radius of convergence $R$ if and only if the largest disc where this series converges is $B(z_0,R) = \{z \in \Bbb C: |z-z_0|<R \}$.
Now, in our problem, notice that the given series diverges at the point $\sqrt 2$. Hence option (1) and (2) are false. [Since $|\sqrt 2| < 2$].
We also have, $|\sqrt 2| \le \sqrt 2$ and so option (3) is also false. Hence the correct answer is option (4).
Alternate solution:
The given series is equal to the geometric series $\sum\limits_{n=0}^{\infty}(\frac{z^2}{2})^n$ which converges if and only if $|\frac{z^2}{2}| < 1$ if and only if $|z|<\sqrt 2$.
### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...