How to show that a sum is inclosed by two values / bounds?

Question

Show that this holds for all c>0

$\frac{\pi}{2\sqrt{c}} \le \sum_{n=0}^{\infty} \frac{1}{n^2 + c} \le \frac{\pi}{2\sqrt{c}} + \frac{1}{c}$

I'd really appreciate it if some can tell me if my proof is valid (I checked the answer and they did it in a much complicated way so I felt this might be incorrect even if it feels like it proves something). Anyways here is my take on it:

The first thing I wanna do is use Cauchy's integral test to see if the sum converges. I still don't fully understand why I picked 0 as a lower bound for the integral (I did it because of the sum starts $a_0$, but I'd appreciate if someone explained it for me assuming I'm even correct).

$\int_{0}^{\infty} \frac{1}{(\frac{x}{\sqrt{c}})^2+1}dx = \lim_{R \to \infty}[\frac{arctan(\frac{x}{\sqrt{c}})}{\sqrt{c}}]_{0}^{R} = \lim_{R \to \infty} \frac{arctan(\frac{R}{\sqrt{c}})}{\sqrt{c}} - \frac{arctan(\frac{0}{\sqrt{c}})}{\sqrt{c}} = \frac{\pi}{2\sqrt{c}}$.

Now I know the integral converges and to $\frac{\pi}{2\sqrt{c}}$

According to Cauchy's integral test, the sum and the integral will converge the same way as n goes to infinity, so I assumed I can now compare the bounds in the question with this.

$\frac{\pi}{2\sqrt{c}} \le \sum_{n=0}^{\infty} \frac{1}{n^2 + c} = \frac{\pi}{2\sqrt{c}} \le \frac{\pi}{2\sqrt{c}} + \frac{1}{c}$

So finally, is this an okay proof? Did I assume too much without proving? Is this even correct?

Answer 1

Define for $c > 0$ the function $f_c(x) = (x^2+c)^{-1}$ and let $$S(c) = \sum_{n=0}^\infty f_c(n). \tag{1}$$ Then since $f_c$ is a strictly decreasing function on $x \in [0,\infty)$, it follows that $$S(c) = \int_{x=0}^\infty f_c(\left\lfloor x \right\rfloor ) \, dx \ge \int_{x=0}^\infty f_c(x) \, dx. \tag{2}$$ We can call $(2)$ the inequality arising from left endpoint discretization of $f_c$. Similarly, $$S(c) - \frac{1}{c} = \sum_{n=1}^\infty f_c(n) = \int_{x=0}^\infty f_c(\left\lceil x \right\rceil) \, dx \le \int_{x=0}^\infty f_c(x) \, dx, \tag{3}$$ which is the corresponding right endpoint discretization. The two inequalities combined yield the compound $$I(c) \le S(c) \le I(c) + \frac{1}{c}, \tag{4}$$ where $$\begin{align} I(c) &= \int_{x=0}^\infty \frac{dx}{x^2 + c} \\ &= \frac{1}{\sqrt{c}} \lim_{r \to \infty} \left[\arctan \frac{x}{\sqrt{c}}\right]_{x=0}^r \\ &= \frac{\pi}{2\sqrt{c}}. \tag{5} \end{align}$$ This concludes the proof.

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The main conceptual difference between my proof and yours is that I explicitly use a property of $f_c$ (namely, that it is a strictly decreasing function) to establish upper and lower bounds on the sum $S_c$; you have not done this. By referring to a convergence test (Cauchy), you are not actually making a claim that indicates whether the integral is greater than or less than the sum, and that is the flaw in your proof. They may have the same convergence properties, but this can also be true for a function that is non-monotone, which would pose problems for bounding it above and below by a suitable integral expression.

Answer 2

First of all, the series $~\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}~$ is known to be convergent and equal $~\dfrac{\pi^2}{6}.$

Also, you have that the terms in the series $~\displaystyle \sum_{n=1}^\infty \frac{1}{n^2 + c}$ are each positive. So, that infinite series is strictly increasing, and bounded above by $~\dfrac{\pi^2}{6},~$ and so must be convergent.

Therefore, the series $\displaystyle \sum_{n=0}^\infty \frac{1}{n^2 + c} = \frac{1}{c} + \sum_{n=1}^\infty \frac{1}{n^2 + c}$ must also be a convergent series.

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$\displaystyle \frac{d ~\arctan \left(\frac{x}{k}\right)}{dx} = \frac{1}{1 + \frac{x^2}{k^2}} \times \frac{1}{k} = \frac{k}{k^2 + x^2}.$

Therefore,

$$\int \frac{k ~dx}{x^2 + k^2} = \arctan \left(\frac{x}{k}\right) \implies $$

$$\int \frac{dx}{x^2 + k^2} = \frac{1}{k} \times \arctan \left(\frac{x}{k}\right).$$

Let

$$I = \int_0^\infty \frac{dx}{x^2 + c} = \lim_{b \to \infty} \left[ ~\int_0^b \frac{dx}{x^2 + c} ~\right] \tag1 $$

$$= \frac{1}{\sqrt{c}} \times \lim_{b \to \infty} \left[ ~\arctan \left(\frac{b}{\sqrt{c}}\right) - \arctan \left(\frac{0}{\sqrt{c}}\right) ~\right]$$

$$= \frac{1}{\sqrt{c}} \times \frac{\pi}{2}.$$

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The problem can be completed by :

• Constructing a set of tall rectangles (set $T$) and short rectangles (set $S$).

• Showing that the combined area of the rectangles in set $~T~$ is
  $~\displaystyle \sum_{n=0}^\infty \frac{1}{n^2 + c}.$

• Showing that the combined area of the rectangles in set $~S~$ is
  $~\displaystyle \sum_{n=1}^\infty \frac{1}{n^2 + c}.$

• Showing that the region represented by the tall rectangles completely covers the area under the curve, represented by the integral $I$, defined in (1) above.

• Showing that the region represented by the short rectangles is completely covered by the area under the curve, represented by the integral $I$, defined in (1) above.

Assuming that each of the above bullet points is accomplished, this will imply that

$$\sum_{n=0}^\infty \frac{1}{n^2 + c} > \frac{\pi}{2\sqrt{c}} > \sum_{n=1}^\infty \frac{1}{n^2 + c} \implies $$

$$\frac{1}{c} + \frac{\pi}{2\sqrt{c}} > \sum_{n=0}^\infty \frac{1}{n^2 + c}.$$

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Note that the integrand represented in (1) above, $~\dfrac{1}{x^2 + c}~$ is always positive, and is strictly decreasing, as $x$ goes from $~0 \to \infty.$

Note further, that for each $~k \in \Bbb{Z_{\geq 0}}$, the height of the integrand at $~x=k~$ is $~\dfrac{1}{k^2 + c}.$

Consider the region under the curve, of the integrand in (1) above, as $~x~$ goes from $~k~$ to $~k + 1$. This region is completely covered by the tall rectangle whose vertices are

$$\left(k,0\right), ~(k+1,0), ~\left(k, ~\left[\frac{1}{k^2 + c} \right] ~\right), ~\left(k+1, ~\left[\frac{1}{k^2 + c} \right] ~\right),$$

and this tall rectangle has an area of $~\displaystyle \frac{1}{k^2 + c}.$

Similarly, as $~x~$ goes from $~k~$ to $~k+1,~$ the area under the curve completely covers the short rectangle whose vertices are

$$\left(k,0\right), ~(k+1,0), ~\left(k, ~\left[\frac{1}{[k+1]^2 + c} \right] ~\right), ~\left(k+1, ~\left[\frac{1}{[k+1]^2 + c} \right] ~\right),$$

and this short rectangle has an area of $\displaystyle \frac{1}{[k+1]^2 + c}.$

So, when you consider the combined area of all of the tall rectangles vs the area under the curve vs the combined area of the short rectangles, the bullet point tasks of the previous section are accomplished.