Question

The interval of convergence of the series \(\displaystyle\sum_{n=0}^\infty x^{(n^2)}\) is:
1. R - [-1, 1]
2. [-1, 1]
3. (-1, 1)
4. R - {0}

Answer 1

Correct Answer - Option 3 : (-1, 1)

Concept :

In mathematics, the radius of convergence of a power series is the radius of the largest disk in which the series converges.

It is either a non-negative real number or ∞.

The radius of convergence can be obtained by applying the ratio test to the terms of the series.

In the ratio test, we generally obtain the value of the limit  \(L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\)

if L < 1 then the series converges, but if L > 1 the series diverges.

Calculation:

Given power series is \(\displaystyle\sum_{n=0}^\infty x^{(n^2)}\)

Let, \(a_n= x^{(n^2)}\)

Then, \(a_{n+1} = x^{((n+1)^2)} = x^{(n^2 + 2n+1)}\)

Now from the ratio test,

\( L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| \)

\( L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^{(n^2+2n+1)}}}}{{{x^{(n^2)}}}}} \right|\)

\(L = \mathop {\lim }\limits_{n \to \infty } \left| {x^{2n+1}} \right|\)

\(= |x| \mathop {\lim }\limits_{n \to \infty } \left| {x^{2n}} \right|\)

\( = |x| \mathop {\lim }\limits_{n \to \infty } \left| {({x^2})^{n}} \right|\)

Let x² = y,

\(L = |x| \mathop {\lim }\limits_{n \to \infty } \left| {{y}^{n}} \right|\)

The given series is convergent (L<0) if;

\(\mathop {\lim }\limits_{n \to \infty } y^n = 0\)

It is only possible when |y| < 1

and otherwise, a limit does not exist.

So for the limit L to exist, |y| < 1 i.e.

|x²| < 1 ⇒ -1 < x < 1

So when -1 < x < 1, 

\(\mathop {\lim }\limits_{n \to \infty } y^n = 0 \Rightarrow L = 0\)

If L = 0 (L < 1), the series is convergent.

∴ for -1 < x < 1, the series is convergent.