Can you show the steps followed for proving divergence or convergence
$\sum\limits_{n=1}^{\infty} $ $\sqrt[n]{n+(-1)^n}$
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can i consider (-1)^n negligible at infinity and what test to choose for such series? – jammah Sep 25 '15 at 19:15
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When proving divergence/convergence, it's useful to bound the summands by something simpler. To deal with the $(-1)^n$, can you think of simple lower and upper bounds for $\sqrt[n]{n+(-1)^n}$? – Adina Goldberg Sep 25 '15 at 19:19
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in this case you can indeed skip the $(-1)^n$ part, to see this use the sandwich lemma. You always have to make sure that something is negligible in the limit, it won't work in general – user190080 Sep 25 '15 at 19:30
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1Knowing limit of $\sqrt[n]n$ might help: http://math.stackexchange.com/questions/115822/how-to-show-that-lim-n-to-infty-n-frac1n-1 – Martin Sleziak Oct 05 '15 at 05:38
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This series clearly diverges (or converges to infinity) since the necessary condition, that in $$ \sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\sqrt[n]{n+(-1)^n} $$ the sequence $a_n$ is a zero-sequence, i.e. that $a_n\to 0$, is not fulfilled. Since $$ \sqrt[n]{n-1}\leq\sqrt[n]{n+(-1)^n}\leq\sqrt[n]{n+1} $$ and $$ \lim_{n\to\infty}\sqrt[n]{n-1}=\lim_{n\to\infty}\sqrt[n]{n+1}=1 $$
user190080
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No because
$$ \sqrt[n]{n+(-1)^n}\rightarrow 1\neq 0.$$ You know that necessary condition(but not sufficient) for $\sum\limits_{n=1}^{\infty} a_n<\infty$ is $a_n\rightarrow 0$
R.N
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HINT:
$n-1 \leq n+(-1)^n \leq n+1$
barak manos
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@jammah: Yes. Please note that the $n$th square of each term is larger than $1$, hence the infinite series diverges. – barak manos Sep 25 '15 at 19:19
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@jammah: A positive root of a positive number larger than $1$ is always larger than $1$. – barak manos Sep 25 '15 at 19:22
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after this step i do the nth root for each but isnt the nth root of n-1 <1 and the nth root of n+1 greater than 1 ? – jammah Sep 25 '15 at 19:23
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