Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove with $n \ge 1$:

$$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot4}+\cdots+\frac{n+2}{n(n+1)2^n} = 1 - \frac{1}{(n+1)2^n}$$

First, I prove it for $n=1$: $$\left(\frac{1+2}{1(1+1)2^1} = 1-\frac{1}{(1+1)2^1}\right) \implies \left(\frac{3}{4} = 1- \frac{1}{4}\right) \implies \left(\frac{3}{4} = \frac{3}{4}\right)$$ Which is true.

So I will now assume this: $$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} = 1 - \frac{1}{(n+1)2^n}$$

And I want to prove it for $n+1$, i.e:

$$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}} = 1 - \frac{1}{((n+1)+1)2^{n+1}}$$

This is how I tried to prove it:

$$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}} =$$

$$ 1 - \frac{1}{(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}}$$

Before continuing, I usually like to grab a calculator, give a value to $n$ and evaluate my current expression with the expression I want to reach. If both values are equal, it means I'm doing okay.

So I took $n=5$ and evaluated the expression I want to reach:

$$1 - \frac{1}{(5+2)\cdot2^5+1} = \frac{447}{448}$$ Then, still with $n = 5$ I evaluated my current expression: $$1 - \frac{1}{(5+1)\cdot2^5}+\frac{5+3}{(5+1)(5+1)\cdot2^{5+1}} = \frac{575}{576}$$

So I got $\frac{447}{448}$ for the expression I want to reach and $\frac{575}{576}$ for what I got so far. Something went wrong.

My problem with this is that I haven't done any calculations yet. All my steps so far were rather mechanical - things I always do with mathematical induction.

Maybe I simply evaluated them wrongly. But I can't see it - I've tested it many times already.

Why are both expressions different? They should be the same.

share|cite|improve this question
In the last term of the last displayed equation, looks like one of those $5+1$s should be a $5+2$. – Gerry Myerson Nov 27 '12 at 6:09
The second is $\dfrac{447}{448}$ after Gerry's correction. – Jonas Meyer Nov 27 '12 at 6:13
@GerryMyerson: You're right. Thank you. Do post the answer :) – Zol Tun Kul Nov 27 '12 at 6:14
why is the denominator in the second term $2\cdot 3\cdot 2$? It should be $2\cdot3 \cdot 4$. – André Nicolas Nov 27 '12 at 6:15
@AndréNicolas: Ah! You're right! Fixing now. – Zol Tun Kul Nov 27 '12 at 6:18
up vote 1 down vote accepted

Elevating comment to answer, at suggestion of OP:

In the last term of the last displayed equation, there is a $5+1$ where there should be a $5+2$. Jonas Meyer notes that this correction gets rid of the discrepancy.

share|cite|improve this answer
Confirmed - it does work now. Thank you. – Zol Tun Kul Nov 27 '12 at 6:22

There is a flaw in the logic in this posting:

begin quote:

First, I prove it for $n=1$: $$\left(\frac{1+2}{1(1+1)2^1} = 1-\frac{1}{(1+1)2^1}\right) \implies \left(\frac{3}{4} = 1- \frac{1}{4}\right) \implies \left(\frac{3}{4} = \frac{3}{4}\right)$$ Which is true.

end of quote

This is not valid reasoning. You're saying "Let's prove $A$, as follows: If $A$ then $3/4 = 3/4$, which is true." Very well then, I shall prove that Socrates was John F. Kennedy: If Socrates was John F. Kennedy, then Socrates died in 1963, and hence, Socrates was mortal. Which is true." To say "If B then A. And A is true. Therefore B." is a standard erroneous form of reasoning. Writers on logic since the time of Aristotle have identified this as a fallacy. A correct way of reasoning is this: $$ \frac{1+2}{1(1+1)2^1} = \frac34 = 1 - \frac14 = 1-\frac{1}{(1+1)2^1}. $$ Therefore the proposition in case $n=1$.

And then you repeat the same fallacy in your inductive step.

You should write "$=$" between things ALREADY KNOWN TO BE EQUAL. Then when you write $$ A = B = C = \cdots = Z, $$ then that proves that $A=Z$. You should not write $$ A = Z $$ therefore $$ B=Y $$ therefore $$ C=X $$ etc. etc. etc. etc.
Therefore $$ 3=3 $$ which is true. Therefore $A=Z$.

share|cite|improve this answer
Ah! You're right about the $n=1$ case. But I'm a bit confused about the inductive step case. I mean, since I am assuming that $\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} = 1 - \frac{1}{(n+1)2^n}$ then I do know that $\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}} = 1 - \frac{1}{(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}}$, I guess. – Zol Tun Kul Nov 27 '12 at 6:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.