大师作文网数列与不等式求证：n＜(k=1,n)∑√(1+(1/k²)+(1/(k+1)²))＜n+1.(n∈N
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数列与不等式
求证：n＜(k=1,n)∑√(1+(1/k²)+(1/(k+1)²))＜n+1.(n∈N+)
求证：n＜(k=1,n)∑√(1+(1/k²)+(1/(k+1)²))＜n+1.(n∈N+)
1+ 1/k²+ 1/(k+1)²
=[k²(k+1)²+(k+1)²+k²]/[k²(k+1)²]
=[k²(k+1)²+2k²+2k+1]/[k²(k+1)²]
=[k²(k+1)²+2k(k+1)+1]/[k²(k+1)²]
=[k(k+1)+1]²/[k(k+1)]²
√[1+ 1/k²+ 1/(k+1)²]
=√[k(k+1)+1]²/[k(k+1)]²
=[k(k+1)+1]/[k(k+1)]
=1+ 1/[k(k+1)]
=1+ 1/k -1/(k+1) /关键就是化简,剩下的就非常简单了.
n
∑ √[1+ 1/k²+ 1/(k+1)²]
k=1
=1+1-1/2+1+1/2-1/3+...+1+1/n -1/(n+1)
=n+[1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=n+1 -1/(n+1)
1/(n+1)>0 n+1- 1/(n+1)n
k=1
综上,得
n
n< ∑ √[1+ 1/k²+ 1/(k+1)²]
=[k²(k+1)²+(k+1)²+k²]/[k²(k+1)²]
=[k²(k+1)²+2k²+2k+1]/[k²(k+1)²]
=[k²(k+1)²+2k(k+1)+1]/[k²(k+1)²]
=[k(k+1)+1]²/[k(k+1)]²
√[1+ 1/k²+ 1/(k+1)²]
=√[k(k+1)+1]²/[k(k+1)]²
=[k(k+1)+1]/[k(k+1)]
=1+ 1/[k(k+1)]
=1+ 1/k -1/(k+1) /关键就是化简,剩下的就非常简单了.
n
∑ √[1+ 1/k²+ 1/(k+1)²]
k=1
=1+1-1/2+1+1/2-1/3+...+1+1/n -1/(n+1)
=n+[1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=n+1 -1/(n+1)
1/(n+1)>0 n+1- 1/(n+1)n
k=1
综上,得
n
n< ∑ √[1+ 1/k²+ 1/(k+1)²]
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