Finding the sum of a sequence of terms

Question

$$1/1(2) - 1/3(2^3) + 1/5(2^5) - 1/7(2^7)$$ This is equal to $$\sum_{n=0}^\infty(1/2)^{2n+1}(-1)^n/(2n+1)$$ Differentiating this leads to: $$\sum_{n=0}^\infty(-1/4)^n$$ Which is equal to $4/5$

Thus, the sum of the original series is equal to: $$\int4/5dx = 4x/5$$

Could some one please confirm that my logic makes sense, specifically when I bring up the point of differentiating the series?

Answer 1

No, partly because your answer contains a variable when it should be a constant.

Answer 2

The original series did not have $x$ anywhere in it, so why should its sum be a function of $x$?

You kinda had the right idea regarding differentiating a series.

Let $f(x) = \displaystyle\sum_{n = 0}^{\infty}\dfrac{(-1)^nx^{2n+1}}{2n+1}$. Then, $f'(x) = \displaystyle\sum_{n = 0}^{\infty}(-1)^nx^{2n} = \dfrac{1}{1+x^2}$.

Since $f(0) = 0$, we have $f(t) = \displaystyle\int_{0}^{t}\dfrac{1}{1+x^2}\,dx$ (which I'll let you evaluate).

Then, your desired sum is simply $f(\tfrac{1}{2})$.

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large\sum_{n=0}^{\infty}% \pars{\half}^{2n+1}\pars{-1}^{n}/\pars{2n+1}} =\sum_{n=0}^{\infty}\pars{\half}^{2n+1}\pars{-1}^{n}\int_{0}^{1}x^{2n}\,\dd x \\[3mm]&=\half\int_{0}^{1}\sum_{n=0}^{\infty}\pars{-\,{x^{2} \over 4\phantom{^{2}}}}^{n}\,\dd x =\half\int_{0}^{1}{\dd x \over 1 - \pars{-x^{2}/4}} =\int_{0}^{1/2}{\dd x \over x^{2} + 1}=\color{#66f}{\large\arctan\pars{\half}} \\[3mm]&\approx {\tt 0.4636} \end{align}