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If the partial sums${\mathbf{\text{S}}}_{\mathbf{n}}$ of an infinite series tend to a limit S, the series is called convergent. If the partial sums${\mathbf{\text{S}}}_{\mathbf{n}}$ of an infinite series don't approach a limit, the series is called divergent.

The limiting value S is called the sum of the series.

The given series is $鈭�\frac{1}{n^2}$.

Use the integral in the given series, $\underset{0}{\overset{鈭�}{鈭�}}{n}^{-2}鈥�dn={\left.-n^{-1}\right|}_0^{鈭�}=0+鈭�=鈭�$.

A student evaluates this integral and concludes that the series diverges. But it is wrong.

Here the upper limit on the integrals is$鈭�$ and the lower limit could be made to correspond to any term of the series. If the integral is finite, then the sum of the series from lower limit is finite, the series converges. On the other hand, if the integral is infinite, then the sum of the series from the lower limit is infinite and the series diverges. Since the beginning terms are of no interest, so simply evaluate the integral, $\underset{}{\overset{鈭�}{鈭�}}{n}^{-2}鈥�dn$.

Solve the integral is as follows:

$\begin{array}{rcl}\underset{}{\overset{鈭�}{鈭�}}{n}^{-2}鈥�dn& =& {\left[\frac{n^{-2+1}}{-2+1}\right]}_{}^{鈭�}\\ & =& -{\left[n^{-1}\right]}^{鈭�}\\ & =& -{\left[\frac{1}{n}\right]}^{鈭�}\\ & =& -\left[\frac{1}{鈭�}\right]\\ & =& 0\end{array}$

Therefore the integral value is a finite and the series converges when the integral is to be evaluated only on upper limit.