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CSIR UGC NET June 2019 Part B Question 23
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Real Analysis
CSIR UGC NET June 2019 Part B Question 23
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We will answer each part one by one.
The given series is an alternating series of the from $\sum (-1)^na_n$ with $a_n=\frac 1n$ is a decreasing sequence that converges to zero. This by alternating series test it is convergent. The given statement is
False.
It is a well-known harmonic series that is divergent. It is also
False.
We know that $\sum \frac{1}{n^2}$ is convergent. Let us assume $\sum \frac{1}{n^2}=k>0$, for some k. For a fixed $m\in\mathbb N$ we have $$\sum_{n=1}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} \frac{1}{(n+m)^2}= k$$ Therefore, we have $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{(n+m)^2}=k\sum_{m=1}^{\infty} 1$$Therefore the given series is divergent. It is also a
False
statement.
We already answered in part 3 that it is
true.
The correct option is
only 4
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Junky
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Very nice explanation. Keep it up.
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