Estimating $\sum_{n=1}^\infty \frac{1}{(2n+1)^2}$ with integrals I have question about integration, i saw this question a lot but i never find the solution of this question. could anyone help me ?
Use integration to find upper and lower bounds that differ by at most 0.1 for the
following sum. (You may need to add the first few terms explicitly and then use integrals to bound the sum of the remaining terms.)
   $\sum_{n=1}^\infty \frac{1}{(2n+1)^2}$
 A: Since $f(x) = \frac1{(2x+1)^2}$ is decreasing and converges, you have the following inequalities:
$$\int_{m}^\infty f(x) \text dx \le \sum_m^\infty f(x) \le \int_{m-1}^\infty f(x) \text dx$$
for any $m \in \mathbb Z^+$. (Do you see why?)
If you choose $m=1$, the integrals on the left and right will be too loose, so you can skip ahead a few terms. So, for example,
$$f(1) + \int_{2}^\infty f(x) \text dx \le f(1) + \sum_2^\infty f(x) \le f(1) + \int_1^\infty f(x) \text dx$$
A: We may exploit creative telescoping:
$$\begin{eqnarray*}\sum_{n\geq 1}\frac{1}{(2n+1)^2}&=&\sum_{n\geq 1}\frac{1}{4n(n+1)}-\sum_{n\geq 1}\frac{1}{4n(n+1)(2n+1)^2}\\&=&\frac{1}{4}-\frac{1}{48}\sum_{n\geq 1}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)+\color{blue}{\frac{1}{48}\sum_{n\geq 1}\frac{7n^2+7n+7}{48 n^3(n+1)^3(2n+1)^2}}\end{eqnarray*}$$
The blue term is pretty small in magnitude: since $\frac{7n^2+7n+7}{n^3(n+1)^3(2n+1)^2}\leq g(n)-g(n+1)$, with $g(n)=\frac{7}{240 n^5}$, we have:
$$ \sum_{n\geq 1}\frac{1}{(2n+1)^2} = \frac{11}{48}+E,\qquad 0\leq E\leq \frac{7}{240}.$$
By creative telescoping we also have:
$$ \sum_{n\geq 1}\frac{1}{(2n+1)^2} = -1+\sum_{m\geq 1}\frac{9}{4m^2\binom{2m}{m}}$$
and the new series converges way faster than the original one, leading to:
$$ \sum_{n\geq 1}\frac{1}{(2n+1)^2} =\frac{209}{896}+E,\quad 0\leq E\leq\frac{1}{2000}.$$
A: 
SIMPLE APPROXIMATION:

We can use the results from THIS, to see that
$$\int_N^\infty \frac{1}{(2x+1)^2}\,dx\le \sum_{n=N}^\infty \frac{1}{(2n+1)^2}\le \frac{1}{(2N+1)^2}+\int_N^\infty \frac{1}{(2x+1)^2}\,dx \tag 1$$
which after carrying out the integral in $(1)$ yields
$$\frac{1}{2(2N+1)}\le  \sum_{n=N}^\infty \frac{1}{(2n+1)^2}\le \frac{1}{(2N+1)^2}+\frac{1}{2(2N+1)} \tag 2$$
In order for $\frac{1}{(2N+1)^2}\le 0.1$, we need $N\ge 2$.  Using $N=2$ in $(2)$, we obtain

$$\bbox[5px,border:2px solid #C0A000]{\frac{19}{90}\le \sum_{n=1}^\infty \frac{1}{(2n+1)^2}\le \frac{19}{90}+ \frac1{25}}$$



EXACT CALCULATION

However, we can easily evaluate the sum by writing
$$\begin{align}
\sum_{n=1}^\infty \frac{1}{n^2}&=1+\sum_{n=1}^\infty \left(\frac{1}{(2n+1)^2}+\frac1{(2n)^2}\right)\\\\
&=1+\sum_{n=1}^\infty \frac{1}{(2n+1)^2}+\frac14 \sum_{n=1}^\infty\frac{1}{n^2}\\\\
\sum_{n=1}^\infty \frac{1}{(2n+1)^2}&=-1+\frac34 \sum_{n=1}^\infty\frac{1}{n^2}\\\\
&=-1+\frac{\pi^2}{8}
\end{align}$$
where we used the well-known result $\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$.


ALTERNATIVE METHODOLOGY $1a$ FOR EXACT EVALUATION
In THIS ANSWER, I presented a way forward that does not require prior knowledge of the value of the series $\displaystyle \sum_{n=1}\frac{1}{n^2}=\frac{\pi^2}{6}$, the Riemann-Zeta Function, or dilogarithm function.  Rather, I proceeded with straightforward integral-based analysis that included application of the residue theorem.



ALTERNATIVE METHODOLOGY $1b$ FOR EXACT EVALUATION

And finally, in THIS ANSWER, I presented a way forward to evaluate the series $\displaystyle \sum_{n=1}\frac{1}{n^2}=\frac{\pi^2}{6}$, without appealing to the Riemann-Zeta Function, the dilogarithm function, or Fourier series.  Rather, I proceeded with an approach the relies on real analysis integration methodologies only, including transformation of coordinates.
A: Bounding with Integrals
Naively, we can bound with
$$
\frac16=\int_1^\infty\frac{\mathrm{d}x}{(2x+1)^2}\le\sum_{k=1}^\infty\frac1{(2n+1)^2}\le\int_0^\infty\frac{\mathrm{d}x}{(2x+1)^2}=\frac12
$$
Or we can start further out
$$
\frac1{14}=\int_3^\infty\frac{\mathrm{d}x}{(2x+1)^2}\le\sum_{k=3}^\infty\frac1{(2n+1)^2}\le\int_2^\infty\frac{\mathrm{d}x}{(2x+1)^2}=\frac1{10}
$$
Therefore,
$$
\frac{701}{3150}=\frac19+\frac1{25}+\frac1{14}\le\sum_{k=1}^\infty\frac1{(2n+1)^2}\le\frac19+\frac1{25}+\frac1{10}=\frac{113}{450}
$$

Approximation without Integration
$$
\sum_{n=1}^\infty\frac1{(2n+1)(2n+3)}\le\sum_{n=1}^\infty\frac1{(2n+1)^2}\le\sum_{n=1}^\infty\frac1{2n(2n+2)}
$$
The sum on the left is
$$
\frac12\sum_{n=1}^\infty\left(\frac1{2n+1}-\frac1{2n+3}\right)=\frac16
$$
easily evaluated since it is a Telescoping Series.
The sum on the right is
$$
\frac12\sum_{n=1}^\infty\left(\frac1{2n}-\frac1{2n+2}\right)=\frac14
$$
which is also a telescoping series.
Therefore,
$$
\frac16\le\sum_{n=1}^\infty\frac1{(2n+1)^2}\le\frac14 
$$
A: Use the geometric series:
$$\frac1{1-r}=\sum_{n=0}^\infty r^n$$
$$\frac1{1-r^2}=\sum_{n=0}^\infty r^{2n}$$
$$\frac r{1-r^2}=\sum_{n=0}^\infty r^{2n+1}$$
$$\frac{\ln(1-x^2)}{-2x}=\frac1x\int_0^x\frac r{1-r^2}\ dr=\frac1x\int_0^x\sum_{n=0}^\infty r^{2n+1}\ dr=\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}$$

$$-\frac12\int_0^1\frac{\ln(1-x^2)}x\ dx=\int_0^1\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}\ dx=\sum_{n=0}^\infty\frac1{(2n+1)^2}$$

Add/subtract first few terms to make this your exact sum.  Then, can you take the error bounds from this?
