Limit of a sequence (2)

The answer is e.

Here is a solution. Let $P(n) =\prod _{ k = 1 }^{ n }{ (1 +\frac { 1 }{ t_{ k } } ) }$

We can write each factor in our product series as

$(1 + \frac { 1 }{ t_{ k } } ) = (\frac { { t }_{ k }+ 1 }{ { t }_{ k } } )= (\frac { { t }_{ k+1 } }{ (k+1) \times { t }_{ k } } )$

As a result we have :

$P(n) =(\frac { { t }_{ 2 } }{ 2 \times { t }_{ 1 } } )\times (\frac { { t }_{ 3 } }{ 3 \times { t }_{ 2 } } )\times ...\times (\frac { { t }_{ n } }{ n \times { t }_{ k-1 } } )\times (\frac { { t }_{ n+1 } }{ (n+1) \times { t }_{ N }\quad } )$

$= \frac { { t }_{ n+1 } }{ n+1)! }$

Now let $T(n) = \sum _{ k = 0 }^{ n }{ \frac { 1 }{ k! } }$

We will show that $T(n) =P(n)$ for all $n \ge 1$ by induction on $n$

Here is our base case for $n = 1$

$T(1) = \frac { 1 }{ 0! } + \frac { 1 }{ 1! } = 2 = (1 + \frac { 1 }{ 1 } ) = P(1)$

Now we'll suppose $T(m) = P(m)$ for some m and use this to show that

$T(m+1) = P(m+1)$ . So

$T(m+1)= T(m) +\frac { 1 }{ (m+1)! }$ , and we have

$P(m+1) = P(m)(1+\frac { 1 }{ { t }_{ m+1 } } )= \frac { { t }_{ m+1 } }{ (m+1)!\quad } (1+\frac { 1 }{ { t }_{ m+1 } } )$

$= \frac { { t }_{ m+1 } }{ (m+1)! } +\frac { 1 }{ (m+1)! }$

$=T(m) +\frac { 1 }{ (m+1)! }= T(m+1)$

Consequently $T(n)$ and $P(n)$ have the same limit as $N\rightarrow \infty$

and since $\lim _{ n\rightarrow \infty }{ T(n) } = e$ we also have :

$\lim _{ n\rightarrow \infty }{ P(n) } = e$

Let me know if there are any mistakes or anything is unclear!