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Beautiful Question of Real analysis, from IIT jam 2018
Question Solution : we discuss options one by one. Recall that a point $p\in\mathbb{R}$ is said to be an interior point of $S\subseteq\mathbb{R}$ if i) $p\in S$ ii) there exists a neighborhood of $p$ which is contained in $S$. Now given that, $2018\in S$ is an interior point of $S$. Hence by definition there exists a neighborhood of $2018$ which is contained in $S$, so that $S$ contains an interval. Hence (A) is true. Now as $S\subseteq\mathbb{R}$, and $2018\in S$ is interior point of $S$ hence, it is an limit point of $S$ and hence there is sequence of points in $S$ which do not converge to $2018$. Hence (B) is also true. Clearly (C) is also true by (A) (since $S$ contains a neighborhood (interval) containing $2018$ and we know by definition every point inside that interval is also an interior point of $S$ ) Finally for (D): we know by definition of an interior point $S$ contains a neighborhood (which is an interval) of point $2018$. Now this neighborhood can be too small such...
Series Question IIT JAM 2018
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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