Let $a_n=\frac{1}{\sqrt{(n^2+1)}}+\frac{1}{\sqrt{(n^2+2)}}+\frac{1}{\sqrt{(n^2+3)}}+\ldots+\frac{1}{\sqrt{(n^2+n)}}$ then will limit of $a_n=0$ Because we know if $a_n$ and $b_n$ are convergent sequences converging to $a$ and $b$ respectively then $a_n+b_n$ converges to $a+b$.
sum of two convergent sequences is also convergent
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sequences-and-series
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0What are the $b_n$? Anyway, $\lim_{n\to\infty}a_n\geq 1$, since $\sum_{i=1}^n\frac{1}{\sqrt{n^2+i}}\geq\frac{n}{\sqrt{(n+1)^2}}$. – 2017-01-31
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0Also you can say that $1/n\geq 1/\sqrt{n^2+i} $ hence a upper bound is 1 – 2017-01-31
1 Answers
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Hold-down. We know it for two-elements sum, and finite by induction, but not for a series. $$a_n=\sum_{1=1}^n\frac{1}{\sqrt{n^2+i}}$$
So in this case, we know that if the series converges, the sequence do it also, but in the other direction is false in general