some telescoping series

Notation: $A$ is the sequence $a_1, a_2, a_3$ etc (index starts at $1$ ).

1. $a_k=\fr{k^2}-\fr{(k+1)^2}=\dfrac{2k+1}{k^2(k+1)^2}$
$A$ = 3/4, 5/36, 7/144, 9/400, 11/900, 13/1764…
$b_n=\sum\limits_{k=1}^n a_k=1-\dfrac{1}{(n+1)^2}$

$\begin{align*} a_1&=a_2=1, a_n a_{n+1}=n. \\ A &= 1, 1, 2, 3/2, 8/3, 15/8, 16/5, 35/16, 128/35 \dots \\ b_n&=\fr{a_n}-\fr{a_{n+1}}=\frac{a_{n+1}-a_n}{n} \\ B&=0,1/2,1/2-2/3=-1/6,2/3-3/8=7/24,3/8-8/15=-19/120,\\ &8/15-5/16=53/240\\ c_n&=\sum\limits_{k=1}^n b_k=1-\fr{a_{n+1}} \\ C&=0,1/2,1/3,5/8,7/15\dots \\ c_3&=1/3=1/2-1/6 \\ c_4&=5/8=1/2-1/6+7/24 \\ c_5&=7/15=1/2-1/6+7/24-19/120 \\ c_6&=11/16=1/2-1/6+7/24-19/120+53/240 \end{align*}$

Diagram of $a_n a_{n+1}=n$ :

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3. $A$ = 1, 10, 10, 100, 100, 1000, 1000 etc $=10^{\floor{n/2}}$
So that $a_n a_{n+1}=10^n$
$b_n=\fr{a_n}-\fr{a_{n+1}}=\frac{a_{n+1}-a_n}{10^n}$ .
$B=9/10,0,9/100,0,9/1000$
$c_n=\sum\limits_{k=1}^n b_k=1-\fr{a_{n+1}}=1-\dfrac{1}{10^{\floor{\frac{n+1}{2}}}}$
$C=9/10,9/10,99/100,99/100\dots$

$\begin{align*} A &= 2, 5, 20, 50, 200, 500 \text{ etc, So that } a_n a_{n+1}=10^n \\ b_n&=\fr{a_n}-\fr{a_{n+1}}=\frac{a_{n+1}-a_n}{10^n}. \\ B&=\frac{3}{10}, \frac{15}{100}, \frac{30}{1000}, \frac{150}{10000}, \frac{300}{100000}\dots\\ c_n&=\sum\limits_{k=1}^n b_k=\fr{a_1}-\fr{a^{n+1}}=1/2-\fr{a_{n+1}} \\ C&= \\ c_1&= 1/2-1/5=3/10=0.3=0.5-0.2 \\ c_2&= 1/2-1/20=\frac{3}{10}+\frac{15}{100}=0.45=0.5-0.05 \\ c_3&= 1/2-1/50=\frac{3}{10}+\frac{15}{100}+\frac{30}{1000}=0.48=0.5-0.02 \\ c_4&= 1/2-1/200=\frac{3}{10}+\frac{15}{100}+\frac{30}{1000}+\frac{150}{10000}=0.495=0.5-0.005 \\ c_5&= 1/2-1/500=\frac{3}{10}+\frac{15}{100}+\frac{30}{1000}+\frac{150}{10000}+\frac{300}{100000}=0.498=0.5-0.002 \\ c_\infty&=0.45+0.045+0.0045+0.00045+\dots=0.49999\dots=0.5 \end{align*}$

5. Something slightly different..

$\begin{align*} \sum \fr{k^2}&=\frac{\Big(\frac{(\frac{(1+\frac{1}{2^2})3^2+2^2}{2^2 3^2})4^2+2^2 3^2}{2^2 3^2 4^2}\Big)5^2+2^2 3^2 4^2 }{2^2 3^2 4^2 5^2}\dots \\ &=\limits^? \Big(\big(((\frac{2^2}{1!^2}+1!^2) \frac{3^2}{2!^2}+2!^2)\frac{4^2}{3!^2}+3!^2\big) \frac{5^2}{4!^2}+4!^2\Big)\frac{6^2}{5!^2}\dots \\ &=\limits^? \Big(\big(\frac{((2^2+1!^4) 3^2+1!^2 2!^4)}{1!^2 2!^2}\frac{4^2}{3!^2}+3!^2\big) \frac{5^2}{4!^2}+4!^2\Big)\frac{6^2}{5!^2}\dots \\ &=\limits^? \Big(\big(\frac{((2^2+1!^4) 3^2+1!^2 2!^4)4^2}{1!^2 2!^2 3!^2}+3!^2\big) \frac{5^2}{4!^2}+4!^2\Big)\frac{6^2}{5!^2}\dots \\ &= \Big(\frac{(((2^2+1) 3^2+2!^4)4^2+2!^2 3!^4)5^2+2!^2 3!^2 4!^4}{2!^2 3!^2 4!^2 5!^2} \Big)6^2+\dots \\ &= \Big(\frac{(((2^2+1) 3^2+2^4)4^2+2^6 3^4)5^2+2^8 3^6 4^4}{2^8 3^6 4^4 5^2} \Big)6^2\dots+ \\ &=1725470784/1194393600 \approx \frac{(2.9\dots)^2}{6}\\ b_n&=\frac{n^2 b_{n-1}+(n-1)!^2}{n!^2} \end{align*}$

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some telescoping series

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