Sorry, I missed the denominator. You should really have used LaTeX. One formula isn't a big deal.Imaxx said:
$$\sum_{n=1}^{\infty} \frac{n^2+3n+1}n^2(n+1)^2}$$
$$\sum_{n=1}^{\infty} \frac{n^2+3n+1}{n^2(n+1)^2}$$archaic said:I feel a bit ashamed that there's a mistake in the code block after getting two likes.
Here's the correct version
Sorry for bumping up the post.Code:$$\sum_{n=1}^{\infty} \frac{n^2+3n+1}{n^2(n+1)^2}$$
The convergence of an infinite series can be determined by using various tests such as the ratio test, comparison test, or the integral test. These tests help to determine if the series will approach a finite value or if it will diverge to infinity.
A convergent series is one in which the sum of all the terms approaches a finite value as the number of terms approaches infinity. On the other hand, a divergent series is one in which the sum of all the terms approaches infinity as the number of terms approaches infinity.
The sum of an infinite series can be found by using various methods such as the geometric series formula, telescoping series, or by using mathematical software. However, not all infinite series have a finite sum, so it is important to determine the convergence of the series first.
No, different methods may be required to solve different types of infinite series. For example, the geometric series formula can only be used for geometric series, while the integral test can only be used for certain types of series. It is important to understand the properties of the series before choosing a method to solve it.
There are some shortcuts that can be used for specific types of series, such as the geometric series formula or the telescoping series method. However, for most infinite series, there is no shortcut and it is important to use the appropriate test or method to determine the convergence and sum of the series.