an =\(\frac {n}{n} + 1\)
Substituting n = 1, 2, 3, 4, 5, we obtain
a1 = \(\frac{1}{1} + 1 = \frac{1}{2}\)
a2 = \(\frac{2}{2} + 1 = \frac{2}{3}\)
a3 = \(\frac{3}{3} + 1 = \frac{3}{4}\)
a4 = \(\frac{4}{4} + 1 = \frac{4}{5}\)
a5 = \(\frac{5}{5} + 1 = \frac{5}{6}\)
Therefore, the required terms are \(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, and \frac {5}{6}\).
If $ \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + ... \infty = \frac{\pi^4}{90}, $ $ \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + ... \infty = \alpha, $ $ \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + ... \infty = \beta, $ then $ \frac{\alpha}{\beta} $ is equal to:
The sum $ 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + ... $ upto $ \infty $ terms, is equal to