sum_(n=1)^(oo)(n+1)/(n!) study if it converges or diverces.

Ratio Test Explanation: We will use the ratio test to determine if the series converges or diverges. The ratio test states that for a series sum_(n=1)^(oo)a_n, if the limit as n approaches infinity of |a_(n+1)/a_n| is less than 1, the series converges; if it is greater than 1, the series diverges; and if it is equal to 1, the test is inconclusive.Find General Expression: First, we need to find a general expression for a_n, which is the nth term of the series. In this case, a_n = (n+1)/(n!).Calculate a_(n+1): Next, we find a_(n+1), which is the (n+1)th term of the series. This is a_(n+1) = ((n+1)+1)/((n+1)!) = (n+2)/((n+1)!).Calculate |a_(n+1)/a_n|: Now we calculate the ratio |a_(n+1)/a_n|. This is |((n+2)/((n+1)!))/((n+1)/(n!))|.Simplify the Ratio: Simplify the ratio by multiplying the numerator and denominator by n! which gives us |((n+2)/((n+1)n!))/((n+1)/(n!))| = |(n+2)/(n+1)^2|.Calculate Limit: Now we take the limit as n approaches infinity of |(n+2)/(n+1)^2|. As n goes to infinity, the highest powers of n in the numerator and denominator dominate, so the limit is |1/n|, which approaches 0.Conclusion: Since the limit is 0, which is less than 1, the ratio test tells us that the series converges.

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Math Problems

Algebra 2

Sum of finite series starts from 1

Full solution

\sum_{n=1}^{\infty} \frac{n+1}{n !}

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study if it converges or diverces.

Ratio Test Explanation: We will use the ratio test to determine if the series converges or diverges. The ratio test states that for a series $\sum_{n=1}^{\infty}a_n$ , if the limit as $n$ approaches infinity of $\left|\frac{a_{n+1}}{a_n}\right|$ is less than $1$ , the series converges; if it is greater than $1$ , the series diverges; and if it is equal to $1$ , the test is inconclusive.
Find General Expression: First, we need to find a general expression for $a_n$ , which is the $n$ th term of the series. In this case, $a_n = \frac{n+1}{n!}.$
Calculate $a_{n+1}$ : Next, we find $a_{n+1}$ , which is the $(n+1)$ th term of the series. This is $a_{n+1} = \frac{(n+1)+1}{(n+1)!} = \frac{n+2}{(n+1)!}.$
Calculate $\left|\frac{a_{n+1}}{a_n}\right|$ : Now we calculate the ratio $\left|\frac{a_{n+1}}{a_n}\right|$ . This is $\left|\frac{\frac{n+2}{(n+1)!}}{\frac{n+1}{n!}}\right|$ .
Simplify the Ratio: Simplify the ratio by multiplying the numerator and denominator by $n!$ which gives us $\left|\frac{(n+2)}{((n+1)n!)}/\frac{(n+1)}{(n!)}\right| = \left|\frac{n+2}{(n+1)^2}\right|$ .
Calculate Limit: Now we take the limit as $n$ approaches infinity of $|\frac{n+2}{(n+1)^2}|$ . As $n$ goes to infinity, the highest powers of $n$ in the numerator and denominator dominate, so the limit is $|\frac{1}{n}|$ , which approaches $0$ .
Conclusion: Since the limit is $0$ , which is less than $1$ , the ratio test tells us that the series converges.