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The sequence $(-1)^{n}(1+n^{-1})$ :

a)Converges to $1$.

b)Converges to $-1$.

c)Converges to $1$ and $-1$

d)Converges to neither $1$ or $-1$.

The sequence looks like:

$$-1-\frac{1}{1}+1+\frac{1}{2}-1-\frac{1}{3}+1+\frac{1}{4}....$$

$$\Rightarrow -\frac{1}{1}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}..$$

$$\Rightarrow - ln(1+1)$$

$$\Rightarrow - ln2$$

Therefore, The series Converges to $-ln2$ and hence (d)Converges to neither $1$ or $-1$..

Is this correct? And is there a better way to solve this which i am missing?

Shraddha
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  • Hint. By setting $u_n=(-1)^{n}(1+n^{-1})$, what is $u_{2n}$? what is $u_{2n+1}$? Observe we are dealing with a sequence not a series ... – Olivier Oloa Apr 26 '17 at 08:00
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    It's nonsense. The sequence is $-2, \frac{3}{2}, ..$. They don't get added up. And even if they did you can't just cancel out the $\pm 1$ as you did. – ancient mathematician Apr 26 '17 at 08:01
  • Oh, made a huge mistake adding them up, but even of i could, why couldn't i cancel $\pm 1$ ? – Shraddha Apr 26 '17 at 08:07
  • So , will the answer be c)? – Shraddha Apr 26 '17 at 08:08
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    your third point - coverges to 1 and -1 - does not make sense. If a sequence converges its limit is well defined (unique) – Johannes Apr 26 '17 at 08:16
  • since $1+\frac{1}{n} \rightarrow 1$ as $n \rightarrow \infty$, the sequence $(-1)^n(1+\frac{1}{n})$ behaves like $(-1)^n$, which diverges by oscillating between $-1$ and $1$... – farruhota May 04 '17 at 12:02

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