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

n=1n+1n! \sum_{n=1}^{\infty} \frac{n+1}{n !} \newlinestudy if it converges or diverces.

Full solution

Q. n=1n+1n! \sum_{n=1}^{\infty} \frac{n+1}{n !} \newlinestudy if it converges or diverces.
  1. 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 n=1an\sum_{n=1}^{\infty}a_n, if the limit as nn approaches infinity of an+1an\left|\frac{a_{n+1}}{a_n}\right| is less than 11, the series converges; if it is greater than 11, the series diverges; and if it is equal to 11, the test is inconclusive.
  2. Find General Expression: First, we need to find a general expression for ana_n, which is the nnth term of the series. In this case, an=n+1n!.a_n = \frac{n+1}{n!}.
  3. Calculate an+1a_{n+1}: Next, we find an+1a_{n+1}, which is the (n+1)(n+1)th term of the series. This is an+1=(n+1)+1(n+1)!=n+2(n+1)!.a_{n+1} = \frac{(n+1)+1}{(n+1)!} = \frac{n+2}{(n+1)!}.
  4. Calculate an+1an\left|\frac{a_{n+1}}{a_n}\right|: Now we calculate the ratio an+1an\left|\frac{a_{n+1}}{a_n}\right|. This is n+2(n+1)!n+1n!\left|\frac{\frac{n+2}{(n+1)!}}{\frac{n+1}{n!}}\right|.
  5. Simplify the Ratio: Simplify the ratio by multiplying the numerator and denominator by n!n! which gives us (n+2)((n+1)n!)/(n+1)(n!)=n+2(n+1)2\left|\frac{(n+2)}{((n+1)n!)}/\frac{(n+1)}{(n!)}\right| = \left|\frac{n+2}{(n+1)^2}\right|.
  6. Calculate Limit: Now we take the limit as nn approaches infinity of n+2(n+1)2|\frac{n+2}{(n+1)^2}|. As nn goes to infinity, the highest powers of nn in the numerator and denominator dominate, so the limit is 1n|\frac{1}{n}|, which approaches 00.
  7. Conclusion: Since the limit is 00, which is less than 11, the ratio test tells us that the series converges.

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