<p>A sequence ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, … is such that ${{u}_{1}}=1$ and</p>
<p style="text-align: center">${{u}_{n+1}}={{u}_{n}}-\frac{2n+1}{{{n}^{2}}{{\left( n+1 \right)}^{2}}}$, for all $n\ge 1$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Given that ${{u}_{n}}=\frac{1}{{{n}^{2}}}$, find $\sum\limits_{n=1}^{N}{\frac{2n+1}{{{n}^{2}}{{\left( n+1 \right)}^{2}}}}$.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Give a reason why the series in part <strong>(i)</strong> is convergent and state the sum to infinity.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Use your answer to part <strong>(i)</strong> to find $\sum\limits_{n=2}^{N}{\frac{2n-1}{{{n}^{2}}{{\left( n-1 \right)}^{2}}}}$.</span></div><p style="text-align: right;">[2]</p></li></ol>

A sequence ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, … is such that ${{u}_{1}}=1$ and

${{u}_{n+1}}={{u}_{n}}-\frac{2n+1}{{{n}^{2}}{{\left( n+1 \right)}^{2}}}$, for all $n\ge 1$.

Given that ${{u}_{n}}=\frac{1}{{{n}^{2}}}$, find $\sum\limits_{n=1}^{N}{\frac{2n+1}{{{n}^{2}}{{\left( n+1 \right)}^{2}}}}$.

Give a reason why the series in part **(i)** is convergent and state the sum to infinity.

Use your answer to part **(i)** to find $\sum\limits_{n=2}^{N}{\frac{2n-1}{{{n}^{2}}{{\left( n-1 \right)}^{2}}}}$.

N2007 P2 Q2 [Modified]

N2007 P2 Q2 [Modified]

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