Sky is the limit! (2)

Calculus Level 5

f ( m ) = ∑ n = 0 m ∏ k = 1 m − 1 ( 1 + n k ) \large f(m) = \sum_{n=0}^{m}\prod_{k=1}^{m-1} \left(1+\frac{n}{k} \right)

For f ( m ) f(m) as defined above, find lim ⁡ m → ∞ f ( m ) m \displaystyle \lim_{m \to \infty} \sqrt [m]{f(m)} .

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The answer is 4.

Rohith M.Athreya by Rohith M.Athreya , 22, India

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1 solution

Sparsh Sarode
Feb 23, 2017

f ( m ) = ∑ n = 0 m ∏ k = 1 m − 1 ( 1 + n k ) \displaystyle f(m)=\sum_{n=0}^{m} \prod_{k=1}^{m-1} \left(1+\dfrac{n}{k} \right)

= ∑ n = 0 m ( ( 1 + n 1 ) ( 1 + n 2 ) . . . ( 1 + n m − 1 ) ) \displaystyle = \sum_{n=0}^{m} \left( \left( 1+\dfrac{n}{1} \right) \left(1+\dfrac{n}{2} \right)... \left(1+\dfrac{n}{m-1} \right) \right)

= 1 + ( 1 + 1 1 ) ( 1 + 1 2 ) . . . ( 1 + 1 m − 1 ) + ( 1 + 2 1 ) ( 1 + 2 2 ) . . . ( 1 + 2 m − 1 ) + . . . + ( 1 + m 1 ) ( 1 + m 2 ) . . . ( 1 + m m − 1 ) = 1+ \left( 1+\dfrac{1}{1} \right) \left(1+\dfrac{1}{2} \right)... \left(1+\dfrac{1}{m-1} \right)+ \left( 1+\dfrac{2}{1} \right) \left(1+\dfrac{2}{2} \right)... \left(1+\dfrac{2}{m-1} \right)+...+ \left( 1+\dfrac{m}{1} \right) \left(1+\dfrac{m}{2} \right)... \left(1+\dfrac{m}{m-1} \right)

= 1 + 2 × 3 × . . . . × m ( m − 1 ) ! + 3 × 4 × . . . × ( m + 1 ) ( m − 1 ) + . . . + ( m + 1 ) ( m + 2 ) . . . ( m + ( m − 1 ) ) ( m − 1 ) =1+\dfrac{2 \times 3 \times.... \times m}{(m-1)!}+\dfrac{3 \times 4 \times... \times (m+1)}{(m-1)}+...+\dfrac{(m+1)(m+2)...(m+(m-1))}{(m-1)}

= 1 + m ! 1 ! ( m − 1 ) ! + ( m + 1 ) ! 2 ! ( m − 1 ) ! + . . . + ( m + ( m − 1 ) ) ! m ! ( m − 1 ) ! =1+\dfrac{m!}{1!(m-1)!}+\dfrac{(m+1)!}{2!(m-1)!}+...+\dfrac{(m+(m-1))!}{m!(m-1)!}

= ( m m ) + ( m m − 1 ) + ( m + 1 m − 1 ) + . . . + ( m + ( m − 1 ) m − 1 ) \displaystyle ={m \choose m} + {m \choose m-1}+{m+1 \choose m-1}+...+{m+(m-1) \choose m-1}

= ( m + 1 m ) + ( m + 1 m − 1 ) + . . . + ( m + ( m − 1 ) m − 1 ) \displaystyle =\ \ \ \ \ \ {m+1 \choose m} \ \ \ + \ \ \ {m+1 \choose m-1} +...+{m+(m-1) \choose m-1}

= ( m + ( m − 1 ) m ) + ( m + ( m − 1 ) m − 1 ) \displaystyle = {m+(m-1) \choose m} + {m+(m-1) \choose m-1}

= ( 2 m m ) \displaystyle = {2m \choose m}

f ( m ) = ( 2 m m ) \displaystyle f(m) = {2m \choose m}

lim ⁡ m → ∞ ( f ( m ) ) 1 m = lim ⁡ m → ∞ ( 2 m m ) 1 m = 4 \displaystyle \lim_{m \rightarrow \infty} (f(m))^{\frac{1}{m}}=\lim_{m \rightarrow \infty} {2m \choose m} ^{\frac{1}{m}} = 4

I will leave it to you to prove the above part or you can check here

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Nice solution.

Hana Wehbi - 4 years, 3 months ago

Did the same

saptarshi dasgupta - 3 years, 2 months ago

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