prove that if 1≤m≤n then (1+1n)m&lt;1+mn+(mn)2

For integer m we can use induction.Indeed, for m=1 we have1+1n&lt;1+1n+1n2Let(1+1n)m&lt;1+mn+m2n2Thus,(1+1n)m+1&lt;(1+mn+m2n2)(1+1n)Thus, it remains to prove that(1+mn+m2n2)(1+1n)&lt;1+m+1n+(m+1)2n2or2m+1n2&gt;m2n3which is true because n≥mFinally, we need to check what happens for m=n:(1+1n)n&lt;1+1+1or(1+1n)n&lt;3which is known.Done!

Answered: "prove that if 1≤m≤n then (1+1n)m<1+mn+(mn)2"

Dexter Odom

Answered question

2022-03-30

prove that if

1 \leq m \leq n

then

{(1 + \frac{1}{n})}^{m} < 1 + \frac{m}{n} + {(\frac{m}{n})}^{2}

Answer & Explanation

armejantm925

Beginner2022-03-31Added 20 answers

For integer m we can use induction.
Indeed, for m=1 we have

1 + \frac{1}{n} < 1 + \frac{1}{n} + \frac{1}{n^{2}}

Let

{(1 + \frac{1}{n})}^{m} < 1 + \frac{m}{n} + \frac{m^{2}}{n^{2}}

Thus,

{(1 + \frac{1}{n})}^{m + 1} < (1 + \frac{m}{n} + \frac{m^{2}}{n^{2}}) (1 + \frac{1}{n})

Thus, it remains to prove that

(1 + \frac{m}{n} + \frac{m^{2}}{n^{2}}) (1 + \frac{1}{n}) < 1 + \frac{m + 1}{n} + \frac{{(m + 1)}^{2}}{n^{2}}

\frac{2 m + 1}{n^{2}} > \frac{m^{2}}{n^{3}}

which is true because

n \geq m

Finally, we need to check what happens for m=n:

{(1 + \frac{1}{n})}^{n} < 1 + 1 + 1

{(1 + \frac{1}{n})}^{n} < 3

which is known.
Done!

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