Square Denominators

For $a=1$ the sum will be $\geq 1$ , and for $a \geq 2$ the sum will be $<\sum_{i=2}^{\infty}\frac{1}{i^2} =\frac{\pi^2}{6}-1<0.9$ . $\boxed{\text{No such expression exists}}$

$\begin{aligned} S & \stackrel{def}{=} \displaystyle \sum_{i=a}^b \frac 1{i^2} \\ & = \sum_{i=1}^b \frac 1{i^2} - \sum_{i=1}^{a-1} \frac 1{i^2} \\ & = \left( \sum_{i=1}^b \frac 1{i^2} \right) + \left( - \sum_{i=1}^{a-1} \frac 1{i^2} \right) \\ & \stackrel{def}{=} \alpha + \beta \end{aligned}$

Let's look at $\alpha$ and $\beta$ individualy.

$\begin{aligned} \alpha & = \sum_{i=1}^b \frac 1{i^2} \\ & < \sum_{i=1}^\infty \frac 1{i^2} \\ & = \frac {\pi^2}{6} \approx 1.64 \end{aligned}$

To work with $\beta$ , let's make the assumption $a \geq 2$ . We will check the case $a = 1$ later.

$\begin{aligned} \beta & = - \sum_{i=1}^{a-1} \frac 1{i^2} \\ & \leq - \sum_{i=1}^1 \frac 1{i^2} \\ & = - 1 \end{aligned}$

If we now combine $\alpha < \frac {\pi^2}{6}$ with $\beta \leq -1$ and plug it into $S = \alpha - \beta$ , we get

$\begin{aligned} S & = \alpha + \beta \\ & < \frac {\pi^2}{6} + (-1) \\ & = \frac {\pi^2}{6} - 1 \approx 0.64 < \frac 9{10} \end{aligned}$

So, no matter how large $b$ is, for $a \geq 2$ , the sum will always be too small.

Now, if $a = 1$

$S_b \stackrel{def}{=} \displaystyle \sum_{i=a=1}^b \frac 1{i^2}$

$S_1 = \frac 1{1^2} = 1 > \frac 9{10}$

$S_{b\geq2} > S_1 > \frac 9{10}$

Now we have covered all cases. $a$ is a positive integer, so it is either 1 or greater than or equal to 2 and in both cases the choice of $b$ can't make the sum equal to $\frac 9{10}$ .