What is wrong with my induction proof?

Question

I have been stuck on this proof for over an hour and just cannot wrap my head around what I am doing wrong.

I have to proof the following: $$P_n:\sum_{i=1}^n\frac i{3^i}= \frac34-\frac {2n+3}{4\cdot3^n} \quad (n \in \mathbb{N})$$

My proof: $$P_0:\; \frac {0}{3^{0}} = 0, \quad \frac {3}{4}-\frac {2\cdot0+3}{4\cdot3^0} = \frac34-\frac34=0.$$

$$P_{n+1}:\; \sum_{i=1}^{n+1}\frac i{3^i} = \frac34-\frac {2(n+1)+3}{4\cdot3^{n+1}}$$

$$\begin{align}P_n \rightarrow P_{n+1}:\;\sum_{i=1}^{n+1}\frac i{3^i}&= \sum_{i=1}^n (\frac i{3^i})+\frac{n+1}{3^{n+1}}\\&=\frac34-\frac {2n+3}{4\cdot3^n}+\frac{n+1}{3^{n+1}}\\&=\frac34-\frac {2n+3}{4\cdot3^n}+\frac{n+1}{3\cdot3^n}\\&=\frac34-\frac {3(2n+3)+4(n+1)}{4\cdot3\cdot3^n}\\&=\frac34-\frac {6n+3+4n+4}{4\cdot3^{n+1}}\\&=\frac34-\frac {10n+7}{4\cdot3^{n+1}}.\end{align}$$

Mistake is here: $\frac {3}{4}-\frac {3(2n+3)\color{red}{-}4(n+1)}{433^{n}}$ (should be minus instead of plus) — Aig, Mar 23 '24 at 13:12
There is a second careless mistake that I pointed out in my red answer :) — Stéphane Jaouen, Mar 23 '24 at 13:19
Since you were talking about formatting difficulties, I took the liberty of making some changes to propose a formatting that seems more relevant to me — Stéphane Jaouen, Mar 23 '24 at 13:26

score 1 · Answer 1 · answered Mar 23 '24 at 13:18

$\sum_{i=1}^{n+1} (\frac {i}{3^{i}}) = \sum_{i=1}^{n} (\frac {i}{3^{i}})\;+\;\frac{n+1}{3^{n+1}}=\\=\frac {3}{4}-\frac {2n+3}{4*3^{n}}+\frac{n+1}{3^{n+1}}=\\=\frac {3}{4}-\frac {2n+3}{4*3^{n}}+\frac{n+1}{3*3^{n}}=\\=\frac {3}{4}-\frac {3*(2n+3)\color{red}-4*(n+1)}{4*3*3^{n}}=\\=\frac {3}{4}-\frac {6n\color{red}{+9}\color{red}-4n\color{red}-4}{4*3^{n+1}}=\\=\frac {3}{4}-\frac {\color{red}2n\color{red}{+5}}{4*3^{n+1}}\\=\frac {3}{4}-\frac {2(n+1)+3}{4*3^{n+1}}$

What is wrong with my induction proof?

1 Answers1