Question

Find the sum of the series \[1+3x+5{{x}^{2}}+7{{x}^{3}}+9{{x}^{4}}+...\text{to infinity}\].

Answer 1

Hint: The given series is a power series. Find out first the radius of convergence of the power series and check whether the series converges or not at infinity.

Complete step by step answer:
The coefficient of each term is a constant, which is independent of x. The given series is of the form \[\sum\limits_{i=0}^{\infty }{{{a}_{i}}{{x}^{i}}}\]. Hence it is a power series where in this case \[{{a}_{0}}\] = 1,\[{{a}_{1}}\] = 3,\[{{a}_{2}}\] = 5 etc. Let the sum of the series is S.
We can notice that the given series can be written as \[\sum\limits_{i=0}^{\infty }{(2i+1){{x}^{i}}}\] where \[{{a}_{i}}=2i+1\].
So, S = \[\sum\limits_{i=0}^{\infty }{(2i+1){{x}^{i}}}\]. We have to find the value of S.
Now as it is a power series it will have a radius of convergence by which we will be able to know for which x values the series converges. We know that if R is the radius of convergence of a power series then \[R=\underset{i\to \infty }{\mathop{\lim }}\,\left| \dfrac{{{a}_{i}}}{{{a}_{i+1}}} \right|\].
So, in this case the radius of convergence is \[R=\underset{i\to \infty }{\mathop{\lim }}\,\left| \dfrac{2i+1}{2i+3} \right|=\underset{i\to \infty }{\mathop{\lim }}\,\dfrac{2i+1}{2i+3}=\underset{i\to \infty }{\mathop{\lim }}\,\dfrac{2+\dfrac{1}{i}}{2+\dfrac{3}{i}}=\dfrac{2}{2}=1\].
So, according to theory the given power series converges if \[\left| x \right|<1\]. It is clear that the series diverges if we put x=1 or x=-1. This can be proved by various convergence tests of infinite series.
So, for all real x if \[\left| x \right|<1\] then the sum S exists finitely.
Now, S = \[\sum\limits_{i=0}^{\infty }{(2i+1){{x}^{i}}}\] = \[\sum\limits_{i=0}^{\infty }{(2i{{x}^{i}}+{{x}^{i}})=}\sum\limits_{i=0}^{\infty }{2i{{x}^{i}}+\sum\limits_{i=0}^{\infty }{{{x}^{i}}}=}2\sum\limits_{i=0}^{\infty }{i{{x}^{i}}+\sum\limits_{i=0}^{\infty }{{{x}^{i}}}}\]
\[\sum\limits_{i=0}^{\infty }{{{x}^{i}}}\] is an infinite geometric progression with first term 1 and common ratio x. So, we know from summation of infinite geometric progression formula that, \[\sum\limits_{i=0}^{\infty }{{{x}^{i}}}=\dfrac{1}{1-x}\]
Let L = \[\sum\limits_{i=0}^{\infty }{i{{x}^{i}}}=x+2{{x}^{2}}+3{{x}^{3}}+4{{x}^{4}}+....\] (1)
So, xL = \[\sum\limits_{i=0}^{\infty }{i{{x}^{i+1}}}={{x}^{2}}+2{{x}^{3}}+3{{x}^{4}}+....\] (2)
Shifting one place towards right and subtracting (2) from (1) we get,
\[L(1-x)=x+{{x}^{2}}+{{x}^{3}}+{{x}^{4}}+....\]
The RHS is equal to \[\sum\limits_{i=1}^{\infty }{{{x}^{i}}}=\dfrac{x}{1-x}\].
So,\[L(1-x)=\dfrac{x}{1-x}\]
Or,\[L=\dfrac{x}{{{(1-x)}^{2}}}\]
So, finally we get, \[S=2L+\dfrac{1}{1-x}=\dfrac{2x}{{{(1-x)}^{2}}}+\dfrac{1}{1-x}\]
\[\Rightarrow \] \[S=\dfrac{1+x}{{{(1-x)}^{2}}}\]
Hence, the sum of the given power series is \[\dfrac{1+x}{{{(1-x)}^{2}}}\] whenever \[\left| x \right|<1\].

Note: The series only converges only when \[\left| x \right|<1\]. For all other values of x, the series does not converge and hence has no such sum.