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sum_(n=1)^(oo)(n+1)/(n!) Study if it converges or diverges using Ratio test.

∑n=1∞n+1n! \sum_{n=1}^{\infty} \frac{n+1}{n !} \newlineStudy if it converges or diverges using Ratio test.

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Q. ∑n=1∞n+1n! \sum_{n=1}^{\infty} \frac{n+1}{n !} \newlineStudy if it converges or diverges using Ratio test.
  1. Write general term: Write down the general term of the series. an=n+1n!a_n = \frac{n+1}{n!}
  2. Apply Ratio Test: Apply the Ratio Test to determine convergence or divergence.\newlineWe need to find the limit of ∣an+1/an∣|a_{n+1}/a_n| as nn approaches infinity.
  3. Calculate an+1a_{n+1}: Calculate an+1a_{n+1}.an+1=n+2(n+1)!a_{n+1} = \frac{n+2}{(n+1)!}
  4. Set up ratio: Set up the ratio ∣an+1an∣\left|\frac{a_{n+1}}{a_n}\right|.\left|\frac{a_{n+\(1\)}}{a_n}\right| = \left|\frac{\frac{n+\(2\)}{(n+\(1\))!}}{\frac{n+\(1\)}{n!}}\right|
  5. Simplify the ratio: Simplify the ratio.\(\newline∣an+1an∣=∣(n+2)(n+1)⋅n!((n+1)!)∣\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(n+2)}{(n+1)} \cdot \frac{n!}{((n+1)!)}\right|\newline∣an+1an∣=∣(n+2)(n+1)2∣\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(n+2)}{(n+1)^2}\right|
  6. Take limit: Take the limit of the ratio as nn approaches infinity.lim⁡n→∞∣n+2(n+1)2∣\lim_{n\to\infty} \left|\frac{n+2}{(n+1)^2}\right|
  7. Evaluate limit: Evaluate the limit. \newlinelim⁡n→∞∣n+2(n+1)2∣=lim⁡n→∞∣1n+1∣\lim_{n\to\infty} \left|\frac{n+2}{(n+1)^2}\right| = \lim_{n\to\infty} \left|\frac{1}{n+1}\right|
  8. Limit is 00: The limit is 00 because as nn gets larger, 1(n+1)\frac{1}{(n+1)} approaches 00.lim⁡n→∞∣1(n+1)∣=0\lim_{n\to\infty} \left|\frac{1}{(n+1)}\right| = 0
  9. Ratio Test conclusion: Since the limit is less than 11, the Ratio Test tells us the series converges.

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