# If $u_n= \int_1^ne^{-t^2}dt$, what can be said about $\sum u_n$?

If

$$\displaystyle u_n= \int_1^ne^{-t^2}dt$$ where $$n=1,2,3...$$

Then Which of the following is true

$$1)$$ both the sequence $$u_n$$ and series $$\sum u_n$$ is convergent

$$2)$$ both the sequence $$u_n$$ and series $$\sum u_n$$ is divergent

$$3)$$ The sequence is convergent and $$\sum u_n$$ is divergent

$$4)$$ $$\displaystyle \lim_{n \to \infty}u_n=\frac{2}{e}$$

The solution I tried - I know that $$\displaystyle\int_0^{\infty} e^{-t^2}=\sqrt \pi$$ ,but how to calculate the $$u_2,u_3...$$ I have no idea please help

• how can i apply that theorem here? – honey kumar Jan 11 at 16:28
• If you know that $\int_0^\infty e^{-t^2}dt$ is finite, then you know if the $u_n$ converge. If the $u_n$ converge to a non-zero value, then you know if the sum in divergent. – irchans Jan 11 at 16:30

The sequence $$u_n$$ is convergent to $$C=\int_{1}^{+\infty}e^{-t^2}\,dt$$. Because of this, the series $$\sum u_n$$ is divergent.

No quantitative estimation is really needed, but if you want one

$$u_n = C-\int_{n}^{+\infty}e^{-t^2}\,dt = C-\frac{1}{2}\int_{n^2}^{+\infty}\frac{dt}{e^t\sqrt{t}}\geq C-\frac{1}{2n}\int_{n^2}^{+\infty}\frac{dt}{e^t}=C-\frac{1}{2ne^{n^2}}.$$

• how $\frac{\sqrt \pi}{2}$? – honey kumar Jan 11 at 16:30
• @TheStudent: Initially I incorrectly assumed the left endpoint of the integration range to be zero, now fixed. – Jack D'Aurizio Jan 11 at 16:31
• i have seen some of your answers ,The magic of inequalities you use to solve question is something beyond my mind , But +1 for your solution. – honey kumar Jan 11 at 16:37

1. The sequence is convergent, because we have $$\lim_{n \rightarrow \infty} \left(\int_{1}^{n} e^{-t^2}\ dt\right) = \int_{1}^{\infty} e^{-t^2} dt = \left(\int_{0}^{\infty} e^{-t^2} dt\right) - \left(\int_{0}^{1} e^{t^2}\ dt\right)$$ and the left hand integral is $$\sqrt{\pi}$$, while the right-hand one must be a finite value(*) because $$t \mapsto e^{-t^2}$$ has no singularities for any $$t \in [0, 1]$$. Hence the limit exists by virtue of this expression thus having a well-defined finite value.
2. The series is divergent. This follows from the fact the above limit is nonzero, as it is a basic theorem that a necessary (but not sufficient) condition for convergence of a series is that the sequence of terms involved must converge to zero at infinity. The fact the limit is nonzero, in turn, is because in $$\int_{1}^{\infty} e^{-t^2}\ dt$$ the integrand $$t \mapsto e^{-t^2}$$ is always positive in $$[1, \infty)$$ (and anywhere else, for that matter).
(*) For what it's worth, the exact value is $$\frac{\sqrt{\pi}}{2} \left(1 - \mathrm{erf}(1)\right) \approx 0.7468$$ (though one might object this is a bit "circular" in a sense, but...).