7. Zadatak

Prvo rješenje.

$$\begin{aligned} &\log\left(1+\frac11\right)+2\log\left(1+\frac12\right)+3\log\left(1+\frac13\right)+\cdots+n\log\left(1+\frac1n\right)\\ &=\log\frac21+2\log\frac32+3\log\frac43+\cdots+n\log\frac{n+1}{n}\\ &=\log2−\log1+2\log3−2\log2+3\log4−3\log3+\cdots+n\log(n+1)−n\log n\\ &=n\log(n+1)−\log2−\log3−\cdots−\log n\\ &=n\log(n+1)−(\log2+\log3+\cdots+\log n)\\ &=n\log(n+1)+\log(n+1)−(\log2+\log3+\cdots+\log n+\log(n+1))\\ &=(n+1)\log(n+1)−\log(2\cdot3\cdot\ldots\cdot n\cdot(n+1))\\ &=(n+1)\log(n+1)−\log((n+1)!). \end{aligned}$$

Drugo rješenje.

$$\begin{aligned} &\log\left[\left(1+\frac11\right)\left(1+\frac12\right)^2\left(1+\frac13\right)^3\cdots\left(1+\frac1n\right)^n\right]\\ &=\log\left[\frac21\cdot\left(\frac32\right)^2\cdot\left(\frac43\right)^3\cdots\left(\frac{n+1}{n}\right)^n\right]\\ &=\log\left[\frac{2\cdot3^2\cdot4^3\cdots(n+1)^n}{1\cdot2^2\cdot3^3\cdots n^n}\right]\\ &=\log\left[\frac11\cdot\frac12\cdot\frac13\cdots\frac1n\cdot(n+1)^n\right]\\ &=\log\left[\frac1{n!}\cdot(n+1)^n\right]\\ &=\log\left[\frac{(n+1)^n\cdot(n+1)}{n!\cdot(n+1)}\right]\\ &=\log\left[\frac{(n+1)^{n+1}}{(n+1)!}\right]\\ &=\log(n+1)^{n+1}−\log((n+1)!)\\ &=(n+1)\log(n+1)−\log((n+1)!). \end{aligned}$$

Treće rješenje.

Dokaz matematičkom indukcijom.

1. Provjera tvrdnje za $n=1$.

$$\log\left(1+\frac11\right)=(1+1)\log(1+1)−\log(2!)\quad\Longrightarrow\quad\log2=2\log2−\log2,$$

što je istinita jednakost.

2. Pretpostavimo da tvrdnja

$$\log\left(1+\frac11\right)+2\log\left(1+\frac12\right)+3\log\left(1+\frac13\right)+\cdots+n\log\left(1+\frac1n\right)=(n+1)\log(n+1)−\log((n+1)!)$$

vrijedi za neki $n\in\mathbb N$ i dokažimo da tada vrijedi i za $n+1$, odnosno da vrijedi

$$\begin{aligned} &\log\left(1+\frac11\right)+2\log\left(1+\frac12\right)+3\log\left(1+\frac13\right)+\cdots+n\log\left(1+\frac1n\right)\\ &\quad +(n+1)\log\left(1+\frac1{n+1}\right)=(n+2)\log(n+2)−\log((n+2)!). \end{aligned}$$

Ako primijenimo pretpostavku na prvih $n$ članova s lijeve strane i dodamo $(n+1)$-vi član, dobivamo redom:

$$\begin{aligned} &(n+1)\log(n+1)−\log((n+1)!)+(n+1)\log\left(1+\frac1{n+1}\right)\\ &=(n+1)\log(n+1)−\log((n+1)!)+(n+1)\log\frac{n+2}{n+1}\\ &=(n+1)\log(n+1)−\log((n+1)!)+(n+1)\log(n+2)−(n+1)\log(n+1)\\ &=(n+1)\log(n+2)−\log((n+1)!)\\ &=(n+1)\log(n+2)+\log(n+2)−\log(n+2)−\log((n+1)!)\\ &=(n+2)\log(n+2)−\log((n+1)!\cdot(n+2))\\ &=(n+2)\log(n+2)−\log((n+2)!). \end{aligned}$$

Time smo dokazali da dana tvrdnja vrijedi za sve prirodne brojeve $n$.