Dave the mischievous astronomer

Calculus Level 3

Dave is very mischievous. He presents Carl a problem:

Take the function $f(n)=\sum^n_{k=0}\log\left(\frac{k+2}{k+3}\right)$ What is $f(-2)$ ?

Carl immediately studies the problem carefully. After some thought he tells Dave that it cannot be done. "You cannot have a smaller upper limit on the summation! This is ludicrous!".

Dave replies with a smile "Ah! But to the eyes of an astronomer, the problem can be solved!".

Can you help Carl solve Dave's problem?

The answer is 0.6931471805599453094172321214581.

1 solution

A Former Brilliant Member
May 12, 2014

Hello! I am Carl's parakeet and I have the solution to the problem!

Firstly, we would like to find an expression for $f(n)$ , rather than have a summation. Write it like this:

$f(n)=\log\frac23+\log\frac34\log\frac45+\cdots\log\frac{n+2}{n+3}$

Then be the rules of logarithms, $\log\frac ab=\log a - \log b$ this allows us to rewrite the sum as $f(n)=\log2-\log3+\log3-\log4+\log4-\log5+\cdots\log(n+2)-\log(n+3)$

Now when I looked at Dave's TELESCOPE I realised that the sum TELESCOPED!

Therefore canceling the terms gives us $f(n)=\log(2)-\log(n+3)$ and simply $f(-2)=\log(2)$ .

I knew that since you are mentioning the term 'astronomer', this must have something to do with a telescopic sum or product. But I am afraid the way you are solving the problem is fallacious. The very statement $f\left( n \right) =\log { \frac { 2 }{ 3 } } +\log { \frac { 3 }{ 4 } } +\log { \frac { 4 }{ 5 } } +\log { \frac { 5 }{ 6 } } +...+\log { \frac { n+2 }{ n+3 } }$ demands $n$ to be a positive integer and hence the summation is not defined.

Kuldeep Guha Mazumder - 5 years, 6 months ago

Dave the mischievous astronomer

The answer is 0.6931471805599453094172321214581.

1 solution

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