The a is Right Where it Should Be

Calculus Level 3

$\displaystyle \sum_{x=1}^{\infty} \left [ \ln\left(\dfrac{x+4}{x}\right)-\ln\left(\dfrac{x+6}{x+2}\right) \right ] =\ln(a)$

Evaluate the following sum above and solve for the value of $a$ .

Try the harder version here

The answer is 15.

1 solution

Chew-Seong Cheong
May 9, 2015

$\displaystyle \sum_{x=1}^\infty {\left[\ln{\left(\frac{x+4}{x}\right)} - \ln{\left( \frac {x+6}{x+2}\right)}\right]} \\ = \ln{\left( \frac{5}{1} \right)} \color{#D61F06} {-\ln{\left(\frac{7}{3} \right)}} + \ln{\left( \frac{6}{2} \right)} \color{#3D99F6} {- \ln{\left(\frac{8}{4} \right)}} \color{#D61F06} {+ \ln{\left( \frac{7}{3} \right)}} \color{#EC7300} {- \ln{\left(\frac{9}{5} \right)}} \\ \quad \color{#3D99F6} {+ \ln{\left( \frac{8}{4} \right)}} \color{#D61F06} { - \ln{\left(\frac{10}{6} \right)} } \color{#EC7300} {+\ln{\left( \frac{9}{5} \right)}} \color{#3D99F6} {-\ln{\left(\frac{11}{7} \right)}} + ... \\ = \ln{\left( \frac{5}{1} \right)} + \ln{\left( \frac{6}{2} \right)} = \ln{5}+\ln{3} = \ln{15} \\ \Rightarrow a = \boxed{15}$

Moderator note:

Yes, telescoping sum is the key here: every other term cancels off in pairs. Simple and elegant. Wonderful!

There's a nice way to present it which uses product notation and logarithm and sum-product formulas.

$S=\sum_{x=1}^\infty\left[\ln\left(\frac{x+4}{x}\right)-\ln\left(\frac{x+6}{x+2}\right)\right]=\sum_{x=1}^\infty\left[\ln\left(\frac{x+4}{x}\cdot\frac{x+2}{x+6}\right)\right]\\ \implies S=\ln\left[\prod_{x=1}^\infty\left(\frac{x+4}{x}\cdot\frac{x+2}{x+6}\right)\right]=\ln\left[\frac{\left(\displaystyle\prod_{x=5}^\infty x\right)\left(\displaystyle\prod_{x=3}^\infty x\right)}{\left(\displaystyle\prod_{x=1}^\infty x\right)\left(\displaystyle\prod_{x=7}^\infty x\right)}\right]\\ \implies S=\ln\left[\frac{5\times 6\times \left(\displaystyle\prod_{x=7}^\infty x\right)\left(\displaystyle\prod_{x=3}^\infty x\right)}{1\times 2\times \left(\displaystyle\prod_{x=3}^\infty x\right)\left(\displaystyle\prod_{x=7}^\infty x\right)}\right]=\ln\left(\frac{30}{2}\right)=\ln(15)$

Prasun Biswas - 6 years, 1 month ago

Almost right. You did it like Chew-Seong Cheong, you should be careful with your notations.

All the products should be in the form of $\displaystyle \lim_{n \to \infty} \prod_{x=a}^n x$ . Else, you would be calculating the arithmetic on infinities, $\ln \left ( \frac{ \infty \times \infty }{\infty \times \infty } \right )$ which is undefined.

Brilliant Mathematics Staff - 6 years, 1 month ago

Right! I agree that there's a bit of abuse of notation in my work. But, don't we generally use the notation $\displaystyle\prod_{x=a}^\infty f(x)$ to denote $\displaystyle\lim_{n\to\infty}\prod_{x=a}^n f(x)$ ?

I used to think that we can use that shorthand notation to denote the limiting product (when upper limit of the product tends to infinity).

Prasun Biswas - 6 years, 1 month ago

@Prasun Biswas – Not necessary. It's like considering the limits of integrals: $\displaystyle \int_{-\infty}^\infty x \, dx$ versus $\displaystyle \lim_{n\to \infty} \int_{-n}^n x \, dx$ . Are they the same thing?

Brilliant Mathematics Staff - 6 years, 1 month ago

@Brilliant Mathematics – Aren't they?

Akiva Weinberger - 6 years, 1 month ago

@Akiva Weinberger – Nope, there's a significant difference precisely because the function $f(x)=x$ is unbounded on $\Bbb{R}$ . As such, the first integral is undefined, whereas in the second integral, you have $n\to\infty$ , i.e., it's a very large finite value that is tending to infinity (but isn't infinity). Since $f(x)=x$ is odd, the second integral has a value of $0$ .

Prasun Biswas - 6 years, 1 month ago

@Prasun Biswas – Ah. So it's the difference between $\displaystyle\lim_{(m,n)\to(-\infty,\infty)}\int_m^nf(x)dx$ and $\displaystyle\lim_{n\to\infty}\int_{-n}^nf(x)dx$ , I guess.

Akiva Weinberger - 6 years, 1 month ago

@Prasun Biswas – Also, while that's a very nice way to think of a limit, it technically doesn't make sense. If $n$ is a finite value, how can it tend to anything (i.e. if $n$ is a constant, how can it change value)? But the technical definition is more confusing, so I guess it's fine.

$\lim_{x\to\infty}f(x)=L$ is defined to mean: "For all $\epsilon>0$ , there exists an $N$ such that for all $n>N$ , we have $\lvert f(n)-L\rvert<\epsilon$ ." Or, alternatively, "For all open sets $O$ containing $L$ , there is an unbounded interval $(N,\infty)$ such that $f((N,\infty))\subseteq O$ ."

Akiva Weinberger - 6 years, 1 month ago

You are right. Edited. Thanks.

Chew-Seong Cheong - 6 years, 1 month ago

The a is Right Where it Should Be

The answer is 15.

1 solution

Moderator note:

0 pending reports