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Csir net june 2018: Real analysis Question (1)

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General Mathematics Question : Given $(x_n)$ is convergent sequence in $\mathbb{R}$ and $(y_n)$ is bounded sequence in $\mathbb{R}$ then we can conclude that, (a) $(x_n+y_n)$ is convergent. (b) $(x_n+y_n)$ is bounded. (c) $(x_n+y_n)$ has no convergent subsequence. (d)$(x_n+y_n)$ has no bounded subsequence. Solution : we discuss options one by one (a) take $x_n=1$ for all $n\in\mathbb{N}$ and $y_n=(-1)^n$ for all $n\in\mathbb{N}$ then, sequence $(x_n)$ is convergent and $(y_n)$ is bounded. But, $(x_n+y_n)=(0,2,0,2,0,2,....)$ which is not convergent. Hence (a) is false. (b) given that sequence $(x_n)$ is convergent and hence it is bounded. Also given that, $(y_n)$ is bounded. So that sum of two bounded sequence $(x_n+y_n)$ is also bounded. Hence (b) is true. (c), (d): By option (b) as sequence $(x_n+y_n)$ is bounded sequence and hence by "Bolzano Weierstrass theorem for sequences", it has convergent (hence bounded) subsequence.  So that, (c),(d) are fa...
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Series Question IIT JAM 2018

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General Mathematics Question :  Solution :   $\sum_{n=1}^\infty (-1)^{n+1}\frac{a_{n+1}}{n!}$ $=\sum_{n=1}^\infty (-1)^{n+1}\frac{n+1+\frac{1}{n+1}}{n!}$ $=\sum_{n=1}^\infty (-1)^{n+1}\frac{n}{n!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n!(n+1)}$ $=\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{(n-1)!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n!}+\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{(n+1)!}$ $=[1-\frac{1}{1!}+\frac{1}{2!}-...]+ [1-\frac{1}{2!}+\frac{1}{3!}-...]+[\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...]$ $=e^{-1}+[1-\frac{1}{2!}+\frac{1}{3!}-...]+[\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...]$ $=e^{-1}+[\mathbf{1-1}+1-\frac{1}{2!}+\frac{1}{3!}-...]+[\mathbf{1-\frac{1}{1!}}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...]$ $=e^{-1}+[1-(1-\frac{1}{1!}+\frac{1}{2!}-...)]+e^{-1}$ $=e^{-1}+[1-e^{-1}]+ e^{-1}$ $=e^{-1}+1$ Hence answer is option (D) Book recommended : Must purchase it nice offer!   For our visitors : Please Support us By...
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