TITLE: $\pi/2=\lim_{y\to{\infty}}\sum_{k=1}^{\infty} 1/ (k(k-1)/y + y/k)$?
QUESTION [0 upvotes]: Algebraic errors over a long period edited multiple times within this body (original post) but basically no major edits although edits critically important.
Main Result:
$\pi/2=\lim_{y\to{\infty}}\sum_{k=1}^{\infty} 1/ (k(k-1)/y + y/k)$.
Givens:
$\phi^{(m)}(z) = (-1)^{(m+1)}m!\sum_{k = 0}^{\infty}1/{(z+k)}^{(m
+1)}$,
and,
$\phi^{(0)}(z+1) = -\alpha + \sum_{k = 1}^{\infty}[z/{(k(k +z))}]$.
Derived from givens:
$\phi^{(0)}(z) = (-1)\sum_{k = 0}^{\infty}[1/{(z+k)}]$.
$\phi^{(0)}(z+1) = (-1)\sum_{k = 0}^{\infty}[1/{(z+1+k)}]$.
$\phi^{(0)}(1) = (-1)\zeta(1)$.
Data:
$y=1E+100$ (to 50 significant digits, keisan online calculator)
$ 2\Phi_0(1/2+iy) =$
$4.6051701859880913680359829093687284152022029772575(E+2) +$
$3.1415926535897932384626433832795028841971693993751(E+0)i$
$\pi =  3.1415926535897932384626433832795028841971693993751$.
The assumption is,
$ \Im[\phi^{(0)}(1/2+iy)] = \pi/2$ when y is sufficiently large.
Derivation of results.
$\Im[\phi^{(0)}(1/2+iy)] = \pi/2$ when y is sufficiently large,
and therefore,
$\pi/2=\lim_{y\to{\infty}}\Im[\phi^{(0)}(1/2+iy)] = \pi/2$.
Let $z+1 = 1/2 + iy$.
Darn, $z = -1/2 + iy$ corrections follow.
Recall,
$\phi^{(0)}(z+1) = -\alpha + \sum_{k = 1}^{\infty}z/{(k(k +z))}$.
Therefore,
$\phi^{(0)}(1/2 + iy) = -\alpha + \sum_{k = 1}^{\infty}[(-1/2 + iy)/{(k(-1/2 + iy + k))}]$.
Taking the imaginary part yields,
$\pi/2= \lim_{y\to\infty}\Im[\sum_{k = 1}^{\infty}(-1/2 + iy)/{(k(-1/2 + iy + k))}]$.
$\pi/2=\lim_{y\to\infty}\Im[\sum_{k = 1}^{\infty} (-1/2 + iy)/(k(k - 1/2 + iy))]$.
Multiplying numerator and denominator by  $(k - 1/2 - iy)$ yields,
$\pi/2 =\lim_{y\to\infty}\Im[\sum_{k = 1}^{\infty} (-1/2 + iy)(k - 1/2 - iy)/(k((k-1/2)^2 + y^2))]$.
Removing the real part yields,
$\pi/2=\lim_{y\to\infty}\Im[\sum_{k = 1}^{\infty}((iy)(k- 1/2) + iy/2)/(k((k-1/2)^2 + y^2))]$.
$\pi/2=\lim_{y\to\infty}\sum_{k = 1}^{\infty}(ky)/(k(k-1/2)^2 + y^2)$.
Rearranging,
$\pi/2=\lim_{y\to\infty}\sum_{k = 1}^{\infty}1/ (k(k-1)/y + 1/(4y)) + y/k)$.
The term $1/(4y)$ drops out and,
$\pi/2=\lim_{y\to\infty}\sum_{k = 1}^{\infty}1/ (k(k-1)/y + y/k)$.
$\pi/2=\lim_{y\to\infty}\sum_{k = 1}^{\infty}k/ (k^2(k-1)/y + y)$.

REPLY [6 votes]: Setting $\delta=1/y$. Then setting $x=n\delta$, we get that $n(n-1)\delta^2=x(x-\delta)$
$$
\begin{align}
\lim_{y\to\infty}\sum_{n=1}^\infty\frac1{n(n-1)/y+y}
&=\lim_{\delta\to0}\sum_{n=1}^\infty\frac1{n(n-1)\delta^2+1}\,\delta\\
&=\int_0^\infty\frac1{x^2+1}\,\mathrm{d}x\\[6pt]
&=\frac\pi2
\end{align}
$$
Since
$$
\underbrace{\sum_{n=1}^\infty\frac1{n^2\delta^2+1}\,\delta}_{\substack{\text{lower Riemann Sum for}\\\int_0^\infty\frac1{x^2+1}\,\mathrm{d}x}}
\le\sum_{n=1}^\infty\frac1{n(n-1)\delta^2+1}\,\delta
\le\underbrace{\sum_{n=1}^\infty\frac1{(n-1)^2\delta^2+1}\,\delta}_{\substack{\text{upper Riemann Sum for}\\\int_0^\infty\frac1{x^2+1}\,\mathrm{d}x}}
$$
{"set_name": "stack_exchange", "score": 0, "question_id": 1670948}