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I would like to show that:

$$ \sum_{k=n+1}^{\infty} e^{-k^2} \sim e^{-(n+1)^2}$$

We have:

$$ \forall p\geq2$$

$$ \exp((n+1)^2-(n+p)^2)=\exp(2n(1-p)+1-p^2)=o(1)$$

$$ e^{-(n+p)^2}=o \left( e^{-(n+1)^2} \right)$$


$$ \sum_{k=n+1}^{\infty} e^{-k^2} =e^{-(n+1)^2}+o \left( e^{-(n+1)^2} \right)\sim e^{-(n+1)^2}$$

Given that there are infinitely many $ o \left(e^{-(n+1)^2} \right)$, can the equality $$ \sum_{k=n+1}^{\infty} e^{-k^2} =e^{-(n+1)^2}+o \left( e^{-(n+1)^2} \right) $$ be directly written?

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You have $n$ as the summing index but not in the expression that is being summed... – Belgi Oct 21 '12 at 11:56
It can't be written this way - summing infinitely many $o(a_n)$ can give any result not neccessary $o(a_n)$ – Norbert Oct 21 '12 at 12:14
up vote 1 down vote accepted

From $e^{-k^2}<\int_{k-1}^k e^{-x^2}dx$ for $k\ge1$ and $\frac d{dx} e^{-x^2}= -2x e^{-x^2}$, we see (for $n>-1$) $$\sum_{k=n+2}^\infty e^{-k^2}< \int_{n+1}^\infty e^{-x^2}\,dx< \frac1{2(n+1)}\int_{n+1}^\infty 2xe^{-x^2}\,dx =\frac{e^{-(n+1)^2}}{2(n+1)},$$ hence $$1<\frac{\sum_{k=n+1}^\infty e^{-k^2}}{e^{-(n+1)^2}}<1+\frac1{2(n+1)}.$$ The estimate is not impressively sharp because we use the very rough estimate $e^{-x^2}<\frac x{n+1}e^{-x^2}$.

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Thank you for this nice answer, I like the inequality with the integrals! – Chon Oct 21 '12 at 14:30

You can sharpen your first identity to $$ \exp((n+1)^2 - (n+p)^2) = \exp(2n(1-p) + 1 - p^2) \le \exp(-2n(p-1)) $$ for all $p \ge 2$. Therefore, $$ 1 \le e^{(n+1)^2} \sum_{j=n+1}^\infty e^{-k^2} \le 1 + \sum_{p=2}^\infty e^{-2n(p-1)} = 1 + \frac{e^{-2n}}{1 - e^{-2n}}\, . $$

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Thank you for your answer Hans Engler! – Chon Oct 21 '12 at 14:29

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