# MA 1505 Tutorial 3: Taylor Series

The Taylor Series of f(x) at the point $x_{0}$ is $f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(x_{0})}{n!} (x-x_{0})^{n}.$ $e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n!}$ $\ln(1+x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{n}}{n}$ $\sin x= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n-1}}{(2n-1)!}$ $\cos x =\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!}$ $\frac{1}{1-x} =\sum_{n=0}^{\infty} x^{n}$

Question 1. Let $S=\sum_{n=0}^{\infty} \frac{1}{n!(n+2)}$. Calculate the value of S.

Solution.

Method (i). $S=\sum_{n=0}^{\infty} \frac{1}{n!(n+2)}$ $= \sum_{n=0}^{\infty} \frac{n+1}{(n+2)!}$ $= \sum_{n=0}^{\infty} \frac{(n+2)-1}{(n+2)!}$ $= \sum_{n=0}^{\infty} (\frac{1}{(n+1)!}-\frac{1}{(n+2)!})$ $= 1$

Method (ii). Integrate the Taylor series of $xe^{x}$ to show that S=1.

The Taylor series of $x e^{x}$ is $\sum_{n=0}^{\infty} \frac{x^{n+1}}{n!}$. Take the integration of the function on the interval [0,1], we get $\int_{0}^{1} xe^{x} dx$ $=\int_{0}^{1} \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} dx$ $= \sum_{n=0}^{\infty} \int_{0}^{\infty} \frac{x^{n+1}}{n!} dx$ $= \sum_{n=0}^{\infty} \frac{1}{n!(n+2)}=S$.

The left hand side equals to 1 from integration by parts.

Method (iii). Differentiate the Taylor series of $(e^{x}-1)/x$.

The Taylor series of $f(x)= (e^{x}-1)/x$ is $\sum_{n=1}^{\infty} \frac{x^{n-1}}{n!}$. Differentiate f(x) and get $f^{'}(x)= \sum_{n=2}^{\infty} \frac{x^{n-2}}{(n-2)!n}$. Moreover, $f^{'}(x)= \frac{e^{x}x-(e^{x}-1)}{x^{2}}$ and $f^{'}(1)=1=S$.

Method (iv). Assume the function $f(x)=\sum_{n=0}^{\infty} x^{n}/(n!(n+2)).$ This implies f(0)=0. Assume $g(x)=\int_{0}^{x}f(t)dt= \sum_{n=0}^{\infty} \int_{0}^{x} \frac{t^{n}}{n!(n+2)}dt = \sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+2)!} = \frac{1}{x} \sum_{n=0}^{\infty} \frac{x^{n+2}}{(n+2)!} = \frac{1}{x}(e^{x}-1-x).$

Since $f(x)=g^{'}(x),$ we get $f(x) = x^{-1}(e^{x}-1)-x^{-2}(e^{x}-1-x).$ That means f(1)=1.

Method (v). Assume the function $f(x)=\sum_{n=0}^{\infty} x^{n}/(n!(n+2)).$ $f(x)= \sum_{n=0}^{\infty} \frac{x^{n}}{(n+1)!} - \frac{x^{n}}{(n+2)!} = x^{-1}\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!} - x^{-2}\sum_{n=0}^{\infty} \frac{x^{n+2}}{(n+2)!} = x^{-1} (e^{x}-1) - x^{-2}(e^{x}-1-x).$ Therefore, f(1)=1.

Remark. There is a similar problem: calculate $\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!(n+2)}.$ Answer is $1-2e^{-1}.$

Question 2. Let n be a positive integer. Prove that $\frac{1}{2} \int_{0}^{1} t^{n-1}(1-t)^{2} dt= \frac{1}{n(n+1)(n+2)}$

and calculate the value of the summation $S=\frac{1}{1\cdot 2 \cdot 3} + \frac{1}{3\cdot 4 \cdot 5} + \frac{1}{5\cdot 6\cdot 7} + \frac{1}{7\cdot 8 \cdot 9}+....$.

Solution. $\frac{1}{2}\int_{0}^{1} t^{n-1}(1-t)^{2}dt$ $= \frac{1}{2} \int_{0}^{1} (t^{n+1}-2t^{n}+t^{n-1}) dt$ $= \frac{1}{2} (\frac{1}{n+2}-\frac{2}{n+1}+\frac{1}{n})$ $= \frac{1}{n(n+1)(n+2)}$.

To calculate the value of S, there are two methods.

Method (i). The summation of S, n is only taken odd numbers. From the first step, we know the summation $S=\frac{1}{2} \int_{0}^{1} (1+t^{2}+t^{4}+t^{6}+...)(1-t)^{2}dt$ $= \frac{1}{2} \int_{0}^{1} \frac{1}{1-t^{2}} (1-t)^{2} dt$ $= \frac{1}{2} \int_{0}^{1} \frac{1-t}{1+t} dt$ $= \frac{1}{2} \int_{0}^{1} (\frac{2}{1+t}-1)dt$ $= \frac{1}{2}( 2\ln(1+t)-t)_{t=0}^{t=1}$ $= \ln 2 -\frac{1}{2}$.

Method (ii).

Since $\frac{1}{n(n+1)(n+2)}= \frac{1}{2}(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2})$, $S=\sum_{ odd} \frac{1}{n(n+1)(n+2)}$ $= \frac{1}{2} \sum_{odd} ( \frac{1}{n}- \frac{2}{n+1}+\frac{1}{n+2})$ $= \frac{1}{2} ( \frac{1}{1}-\frac{2}{2}+\frac{1}{3}+ \frac{1}{3}-\frac{2}{4}+\frac{1}{5}+ \frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{7}-\frac{2}{8}+\frac{1}{9}+...)$ $= \frac{1}{2} ( \frac{1}{1} + 2( -\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-...))$ $= \frac{1}{2} ( 1 + 2 (\ln 2-1))$ $= \ln 2 -\frac{1}{2}$.

Here we use the Taylor series of $\ln(1+x)= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{n}}{n}$ and $\ln 2= 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...$.

Question 3. Assume $\zeta(k)=1+\frac{1}{2^{k}} + \frac{1}{3^{k}} + ... = \sum_{m=1}^{\infty} \frac{1}{m^{k}}.$

Prove $\sum_{k=2}^{\infty} (\zeta(k)-1)=1.$ $\sum_{k=1}^{\infty} (\zeta(2k)-1)=3/4.$

Proof. $\sum_{k=2}^{\infty} (\zeta(k)-1)$ $= \sum_{k=2}^{\infty} \sum_{m=2}^{\infty} \frac{1}{m^{k}}$ $= \sum_{m=2}^{\infty} \sum_{k=2}^{\infty} \frac{1}{m^{k}}$ $= \sum_{m=2}^{\infty} \frac{1}{(m-1)m}$ $= \sum_{m=2}^{\infty} ( \frac{1}{m-1} - \frac{1}{m})$ $= 1.$ $\sum_{k=1}^{\infty} ( \zeta(2k)-1)$ $= \sum_{k=1}^{\infty} \sum_{m=2}^{\infty} \frac{1}{m^{2k}}$ $= \sum_{m=2}^{\infty} \sum_{k=1}^{\infty} \frac{1}{m^{2k}}$ $= \sum_{m=2}^{\infty} \frac{1}{m^{2}-1}$ $= \sum_{m=2}^{\infty} \frac{1}{2} ( \frac{1}{m-1}-\frac{1}{m+1})$ $= \frac{1}{2}(1+\frac{1}{2})$ $= \frac{3}{4}.$

Question 4. Calculate the summation $S= \sum_{k=1}^{\infty} \frac{(-1)^{k} k}{4k^{2}-1}.$

Solution. $S=\sum_{k=1}^{\infty} \frac{(-1)^{k} k}{4k^{2}-1}$ $=\frac{1}{4} \sum_{k=1}^{\infty} (-1)^{k} ( \frac{1}{2k-1} +\frac{1}{2k+1})$ $= \frac{1}{4} \sum_{k=1}^{\infty} ( \frac{(-1)^{k}}{2k-1} - \frac{(-1)^{k+1}}{2k+1})$ $= \frac{1}{4} \cdot \frac{-1}{2-1} = - \frac{1}{4}.$

Advertisements