Question

How do you evaluate an infinite series?

Answer 1

Hint: We recall the definition of convergence infinite series and convergent infinite series. We recall the convergent geometric progression where there is a common ratio between terms, telescoping sums, and special series like Taylor’s series to evaluate sum. \[\]

Complete step-by-step solution:
We know that a sequence is defined as the enumerated collection of numbers where repetitions are allowed and order of the numbers matters. The members of the sequence are called terms. Mathematically, a sequence with infinite terms is written as
\[\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},....\]
We know that the sum of terms in an infinite sequence is called an infinite serie which is given by
\[S={{x}_{1}}+{{x}_{2}}+{{x}_{3}}+...=\sum\limits_{k=1}^{\infty }{{{x}_{k}}}\]
The ${{n}^{\text{th}}}$ partial sum of infinite series is the sum of first $n$ terms that is ${{S}_{n}}=\sum\limits_{k=1}^{n}{{{x}_{k}}}$. A infinite series is said to be convergent when its partial sums ${{S}_{1}},{{S}_{2}},{{S}_{3}},...$ tends to a limit. Mathematically if $l$ is the limit and for arbitrary positive small number $\varepsilon $ there exits ${{n}_{0}}\in \mathsf{\mathbb{N}}$ such that
\[\left| {{S}_{n}}-{{n}_{0}} \right|<\varepsilon \]
 Here $l$ is called the sum of series and We can find the sum of infinite series only when it is convergent. We know that the series of GP(geometric progression ) is a series with where the ratio between consucative constant which means
\[r=\dfrac{{{x}_{2}}}{{{x}_{1}}}=\dfrac{{{x}_{3}}}{{{x}_{2}}}=...\]
Here $r$ is called common ratio and the GP series is convergent only when $\left| r \right|<1$. The sum of convergent GP series with infinite terms with common ratio $r$ is given by
\[S=\dfrac{a}{1-r}\]
We can also evaluate an infinite series using telescoping sums. For example let's evaluate $\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{n}^{2}}+n}}$. We can write the summation as
\[\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{n}^{2}}+n}}=\sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}=}\sum\limits_{n=1}^{\infty }{\left( \dfrac{1}{n}-\dfrac{1}{n+1} \right)}\]
Now we expand the sum as;
\[\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{n}^{2}}+n}}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...=1\]
We see that each term is cancelled out by the previous term, it is called telescoping sum. We can also use Taylor’s series to evaluate an infinite series. We know that a function $f\left( x \right)$ can be expressed into an infinite series for different functions like exponential function ${{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+...$. So let us evaluate $\sum\limits_{n=1}^{\infty }{\dfrac{{{2}^{n}}}{n!}}$. We can write the summation as
\[\begin{align}
  & \sum\limits_{n=1}^{\infty }{\dfrac{{{2}^{n}}}{n!}}=1+\sum\limits_{n=1}^{\infty }{\dfrac{{{2}^{n}}}{n!}}-1 \\
 & \Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{{{2}^{n}}}{n!}}=1+2+\dfrac{{{2}^{2}}}{2!}+\dfrac{{{2}^{3}}}{3!}...-1 \\
 & \Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{{{2}^{n}}}{n!}}={{e}^{2}}-1 \\
\end{align}\]

Note: We note that if a series is not convergent we call the series a divergent series and we cannot find the sum of a divergent series. The Taylor’s series approximation for any function $f\left( x \right)$ that is infinite times differentiable at some point $x=a$ is given by $f\left( x \right)=f\left( a \right)+{{f}^{'}}\left( a \right)\left( x-a \right)+\dfrac{{{f}^{''}}\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+...$ and from where we can derive Maclaurine at $x=0$ series $f\left( x \right)=f\left( 0 \right)+x{{f}^{'}}\left( 0 \right)+{{x}^{2}}\dfrac{{{f}^{''}}\left( a \right)}{2!}+...$. Now we can use $\log \left( 1+x \right),{{e}^{x}},\sin x$ etc to find their series.