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Question:

If f(n) = 1 + 2 + 3 + ... + (n + 1) and g(n) = $\underset{k=1}{\overset{k=n}{\sum \frac{1}{f\left(k\right)}}}$, then the least value of n for which g(n) exceeds the value of $\frac{99}{100}$ is

Explanation:

Given, f(n) = 1 + 2 + 3 + ... + (n+1) = $\frac{(\mathrm{n}+1)(\mathrm{n}+2)}{2}$

Now,

$\underset{k=1}{\overset{k=n}{\sum \frac{1}{f\left(k\right)}}}$ = $\frac{1}{\mathrm{f}\left(1\right)}$ + $\frac{1}{\mathrm{f}\left(2\right)}$ + $\frac{1}{\mathrm{f}\left(3\right)}$ + ... + $\frac{1}{\mathrm{f}\left(\mathrm{n}\right)}$

= $\frac{2}{2\times 3}$ + $\frac{2}{3\times 4}$ + $\frac{2}{4\times 5}$ + ... + $\frac{2}{\left(\mathrm{n}\right)\times (\mathrm{n}+1)}$ + $\frac{2}{(\mathrm{n}+1)\times (\mathrm{n}+2)}$

= $2\left[\frac{1}{2\times 3}+\frac{1}{3\times 4}+\frac{1}{4\times 5}+...+\frac{1}{\mathrm{n}\times (\mathrm{n}+1)}+\frac{1}{(\mathrm{n}+1)\times (\mathrm{n}+2)}\right]$

= $2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}+\frac{1}{(\mathrm{n}+1)}-\frac{1}{\mathrm{n}+2}\right]$

= $2\left[\frac{1}{2}-\frac{1}{\mathrm{n}+2}\right]$

= $2\left[\frac{\mathrm{n}}{2(\mathrm{n}+2)}\right]$

= $\frac{\mathrm{n}}{(\mathrm{n}+2)}$

Now, this should be greater than 99/100

⇒ $\frac{\mathrm{n}}{(\mathrm{n}+2)}$ > $\frac{99}{100}$

⇒ 100n > 99n +198

⇒ n > 198

∴ Least possible integer value of n = 199.

Hence, 199.