For any natural number n, prove the validity of given series by mathematical induction:
2 ( n + 1 − 1 ) < 1 + ( 1 2 ) + . . . + ( 1 n ) < 2 n 2(\sqrt{n+1}-1)<1+(\dfrac{1}{\sqrt{2}})+...+(\dfrac{1}{\sqrt{n}})<2\sqrt{n} 2 ( n + 1 − 1 ) < 1 + ( 2 1 ) + ... + ( n 1 ) < 2 n Let P ( n ) P(n) P ( n ) n n n
2 ( n + 1 − 1 ) < 1 + ( 1 2 ) + . . . + ( 1 n ) < 2 n 2(\sqrt{n+1}-1)<1+(\dfrac{1}{\sqrt{2}})+...+(\dfrac{1}{\sqrt{n}})<2\sqrt{n} 2 ( n + 1 − 1 ) < 1 + ( 2 1 ) + ... + ( n 1 ) < 2 n BASIS STEP: P ( 1 ) P(1) P ( 1 )
2 ( 1 + 1 − 1 ) < 1 < 2 1 2(\sqrt{1+1}-1)<1<2\sqrt{1} 2 ( 1 + 1 − 1 ) < 1 < 2 1
2 ( 2 − 1 ) < 1 < 2 2(\sqrt{2}-1)<1<2 2 ( 2 − 1 ) < 1 < 2 INDUCTIVE STEP: For the inductive hypothesis we assume that P ( k ) P(k) P ( k ) k . k. k .
2 ( k + 1 − 1 ) < 1 + ( 1 2 ) + . . . + ( 1 k ) < 2 k 2(\sqrt{k+1}-1)<1+(\dfrac{1}{\sqrt{2}})+...+(\dfrac{1}{\sqrt{k}})<2\sqrt{k} 2 ( k + 1 − 1 ) < 1 + ( 2 1 ) + ... + ( k 1 ) < 2 k Under this assumption, it must be shown that P ( k + 1 ) P(k+1) P ( k + 1 )
2 ( ( k + 1 ) + 1 − 1 ) < 1 + ( 1 2 ) + . . . + ( 1 k ) 2(\sqrt{(k+1)+1}-1)<1+(\dfrac{1}{\sqrt{2}})+...+(\dfrac{1}{\sqrt{k}}) 2 ( ( k + 1 ) + 1 − 1 ) < 1 + ( 2 1 ) + ... + ( k 1 )
+ ( 1 k + 1 ) < 2 k + 1 +(\dfrac{1}{\sqrt{k+1}})<2\sqrt{k+1} + ( k + 1 1 ) < 2 k + 1 We have that
2 ( k + 1 − 1 ) + ( 1 k + 1 ) < 1 + ( 1 2 ) + . . . + ( 1 k ) + ( 1 k + 1 ) 2(\sqrt{k+1}-1)+(\dfrac{1}{\sqrt{k+1}})<1+(\dfrac{1}{\sqrt{2}})+...+(\dfrac{1}{\sqrt{k}})+(\dfrac{1}{\sqrt{k+1}}) 2 ( k + 1 − 1 ) + ( k + 1 1 ) < 1 + ( 2 1 ) + ... + ( k 1 ) + ( k + 1 1 )
< 2 k + ( 1 k + 1 ) <2\sqrt{k}+(\dfrac{1}{\sqrt{k+1}}) < 2 k + ( k + 1 1 ) Show that
2 ( k + 2 − 1 ) < 2 ( k + 1 − 1 ) + ( 1 k + 1 ) 2(\sqrt{k+2}-1)<2(\sqrt{k+1}-1)+(\dfrac{1}{\sqrt{k+1}}) 2 ( k + 2 − 1 ) < 2 ( k + 1 − 1 ) + ( k + 1 1 )
2 ( k + 2 − k + 1 ) < 1 k + 1 2(\sqrt{k+2}-\sqrt{k+1})<\dfrac{1}{\sqrt{k+1}} 2 ( k + 2 − k + 1 ) < k + 1 1
2 ( k + 2 − k − 1 ) k + 1 + k + 2 < 1 k + 1 \dfrac{2(k+2-k-1)}{\sqrt{k+1}+\sqrt{k+2}}<\dfrac{1}{\sqrt{k+1}} k + 1 + k + 2 2 ( k + 2 − k − 1 ) < k + 1 1
2 k + 1 + k + 2 < 1 k + 1 \dfrac{2}{\sqrt{k+1}+\sqrt{k+2}}<\dfrac{1}{\sqrt{k+1}} k + 1 + k + 2 2 < k + 1 1
2 k + 1 + k + 2 < 2 k + 1 + k + 1 = 1 k + 1 , \dfrac{2}{\sqrt{k+1}+\sqrt{k+2}}<\dfrac{2}{\sqrt{k+1}+\sqrt{k+1}}=\dfrac{1}{\sqrt{k+1}}, k + 1 + k + 2 2 < k + 1 + k + 1 2 = k + 1 1 , True for k ≥ 1. k\geq1. k ≥ 1.
Show that
2 k + ( 1 k + 1 ) < 2 k + 1 2\sqrt{k}+(\dfrac{1}{\sqrt{k+1}})<2\sqrt{k+1} 2 k + ( k + 1 1 ) < 2 k + 1
2 ( k + 1 − k ) k + k + 1 > 1 k + 1 \dfrac{2(k+1-k)}{\sqrt{k}+\sqrt{k+1}}>\dfrac{1}{\sqrt{k+1}} k + k + 1 2 ( k + 1 − k ) > k + 1 1
2 k + k + 1 > 1 k + 1 \dfrac{2}{\sqrt{k}+\sqrt{k+1}}>\dfrac{1}{\sqrt{k+1}} k + k + 1 2 > k + 1 1
2 k + k + 1 > 2 k + 1 + k + 1 = 1 k + 1 \dfrac{2}{\sqrt{k}+\sqrt{k+1}}>\dfrac{2}{\sqrt{k+1}+\sqrt{k+1}}=\dfrac{1}{\sqrt{k+1}} k + k + 1 2 > k + 1 + k + 1 2 = k + 1 1 True for k ≥ 1. k\geq1. k ≥ 1.
We show that P ( k + 1 ) P(k+1) P ( k + 1 ) P ( k ) P(k) P ( k )
We have completed the basis step and the inductive step, so by mathematical induction we know that P ( n ) P(n) P ( n ) n . n. n .
That is, we have proven that for any natural number n , n, n ,
2 ( n + 1 − 1 ) < 1 + ( 1 2 ) + . . . + ( 1 n ) < 2 n 2(\sqrt{n+1}-1)<1+(\dfrac{1}{\sqrt{2}})+...+(\dfrac{1}{\sqrt{n}})<2\sqrt{n} 2 ( n + 1 − 1 ) < 1 + ( 2 1 ) + ... + ( n 1 ) < 2 n 
#339914 on Dec 2023
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