Question

How can I prove that `e^(1/n) <= n/(n-1)` for every `n>=2` ?

Answer 1

`"See explanation"`

Explanation:

`e^x = 1 + x + x^2/2 + x^3/6 + ..." (Taylor series of exp(x))"`
`=> e^(1/n) = 1 + 1/n + (1/n)^2/2 + ...`
`<= 1 + 1/n + (1/n)^2 + ... = 1/(1 - 1/n) = n/(n - 1)" (geometric series )"`
`"(And that geometric series is convergent for n>1)"`

Answer 2

See below.

Explanation:

We are given

`e^(1/n) <= n/(n-1)`

If we take the logarithm of both sides

`ln{e^(1/n)} <= ln{n/(n-1)}`

Simplify using some logarithm properties

`color(blue)(1/n <= ln(n) - ln(n-1))`

This equation should be useful enough, now we just need to definitively show that it is true.

We know for `n >= 2` just by observation and knowledge of logarithms that

`0<1/n <= 1/2 -> "max"{1/n} = 1/2`

`ln(n) - ln(n-1) >= ln(2)->"min"{ln(n)-ln(n-1)} = ln(2)`

Clearly, `1/2 < ln(2)`.

This means `"max"{1/n} < "min"{ln(n)-ln(n-1)}`

This immediately implies that

`=>1/n < ln(n) - ln(n-1)`

which obviously satisfies

`color(blue)(1/n <= ln(n) - ln(n-1))`

which we derived from `e^(1/n) <= n/(n-1)`. Hence

`=>e^(1/n) <= n/(n-1)` is true for `n>=2`