<ol class="i-list"><li><div class="list-item-paragraph">By considering $\mathrm{f}\left( r \right)-\mathrm{f}\left( r+1 \right)$, where&nbsp;$\mathrm{f}\left( r \right)=\frac{1}{r}$, find the sum to $2n$ terms of the series $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...$</div></li><li><div class="list-item-paragraph">Show that the sum to $2n$ terms of the series $\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+\frac{1}{{{5}^{2}}}+...$ is less than $1$.</div></li></ol>

By considering $\mathrm{f}\left( r \right)-\mathrm{f}\left( r+1 \right)$, where $\mathrm{f}\left( r \right)=\frac{1}{r}$, find the sum to $2n$ terms of the series $\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...$

Show that the sum to $2n$ terms of the series $\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+\frac{1}{{{5}^{2}}}+...$ is less than $1$.

Default solution

\[f(r) - f(r+1) = \dfrac{1}{r} - \dfrac{1}{r+1} = \dfrac{r+1-r}{r(r+1)} = \dfrac{1}{r(r+1)}\]

\[\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{20} + \ldots \quad (2n \text{ terms})\]

\[= \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} + \dfrac{1}{4 \cdot 5} + \dfrac{1}{5 \cdot 6} + \ldots\]

\[= \sum_{r=1}^{2n} \dfrac{1}{r(r+1)} = \sum_{r=1}^{2n} \left[\dfrac{1}{r} - \dfrac{1}{r+1}\right]\]

\[= \left[1 - \dfrac{1}{2}\right] + \left[\dfrac{1}{2} - \dfrac{1}{3}\right] + \left[\dfrac{1}{3} - \dfrac{1}{4}\right] + \ldots + \left[\dfrac{1}{2n} - \dfrac{1}{2n+1}\right]\]

\[\dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + \dfrac{1}{5^2} + \ldots = \sum_{r=1}^{2n} \dfrac{1}{(r+1)^2}\]

For $r \geq 1$, $r(r+1) < (r+1)(r+1)$

\[\dfrac{1}{r(r+1)} > \dfrac{1}{(r+1)^2}\]

\[\therefore \sum_{r=1}^{2n} \dfrac{1}{r(r+1)} > \sum_{r=1}^{2n} \dfrac{1}{(r+1)^2}\]

As $n \to \infty$, $\dfrac{1}{2n+1} \to 0$, so $\sum_{r=1}^{2n} \dfrac{1}{r(r+1)} = 1$

\[1 > \sum_{r=1}^{2n} \dfrac{1}{r(r+1)} > \sum_{r=1}^{2n} \dfrac{1}{(r+1)^2}\]

\[\therefore \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + \dfrac{1}{5^2} + \ldots < 1\]

Worked Example 4

Worked Example 4

Video Solution:

Default solution