Let $a_{1}=1$ and for $n \ge 1$,$a_{n+1} = \frac{1}{2}a_{n} + \frac{n^{2}-2n-1}{n^{2}(n+1)^{2}}$. Then $|\sum_{n=1}^{\infty}(a_{n}-\frac{2}{n^{2}})|$ is equal to ........... .

Two sequences $\{t_n\}$ and $\{s_n\}$ are defined by $t_n = \log \left( \frac{5^{n+1}}{3^{n-1}} \right)$ and $s_n = \left[ \log \left( \frac{5}{3} \right) \right]^n$. Then:

If $x=\sum_{n=0}^{\infty} \cos ^{2 n} \theta$,$y=\sum_{n=0}^{\infty} \sin ^{2 n} \theta$,$z=\sum_{n=0}^{\infty} \cos ^{2 n} \theta \sin ^{2 n} \theta$ and $0 < \theta < \frac{\pi}{2}$,then

Consider the sequence $a_{1}, a_{2}, a_{3}, \ldots$ such that $a_{1}=1, a_{2}=2$ and $a_{n+2}=\frac{2}{a_{n+1}}+a_{n}$ for $n=1, 2, 3, \ldots$. If $\left(\frac{a_{1}+\frac{1}{a_{2}}}{a_{3}}\right) \cdot\left(\frac{a_{2}+\frac{1}{a_{3}}}{a_{4}}\right) \cdot\left(\frac{a_{3}+\frac{1}{a_{4}}}{a_{5}}\right) \cdots\left(\frac{a_{30}+\frac{1}{a_{31}}}{a_{32}}\right)=2^{\alpha}\left({}^{61}C_{31}\right)$,then $\alpha$ is equal to.

Let $\{a_k\}$ and $\{b_k\}, k \in N$,be two $G$.$P$.s with common ratios $r_1$ and $r_2$ respectively such that $a_1=b_1=4$ and $r_1 < r_2$. Let $c_k=a_k+b_k, k \in N$. If $c_2=5$ and $c_3=13/4$,then $\sum_{k=1}^{\infty} c_k - (12a_6 + 8b_4)$ is equal to

If $m$ arithmetic means $(A.Ms)$ and three geometric means $(G.Ms)$ are inserted between $3$ and $243$ such that the $4^{\text{th}}$ $A.M.$ is equal to the $2^{\text{nd}}$ $G.M.$,then $m$ is equal to: