I have some difficulty to prove the following limit: limN→∞∑k=1N1k+N=ln⁡(2)

Question

I have some difficulty to prove the following limit:  limN→∞∑k=1N1k+N=ln⁡(2)

Prince Huang · Accepted Answer

A common estimate for the Harmonic Numbers is  ∑k=1n1k=log⁡(n)+γ+O(1n) (1)  where γ is the Euler-Mascheroni constant.  Applying (1), we get that  ∑k=1N1k+N=∑k=12N1k−∑k=1N1k  =(log⁡(2N)+γ+O(12N))−(log⁡(N)+γ+O(1N))  =log⁡(2)+O(1N) (2)  Taking the limit of (2) as N→∞ yields  limN→∞∑k=1N1k+N=log⁡(2)

oferenteoo · Accepted Answer

f(x)=limn→∞∑k=1n1k+nx  You are looking for f(1)  f(x)=limn→∞xn∑k=1n11+kxn=limn→∞xn∑k=1n(1−kxn+k2x2n2−k3x3n3+…)  =limn→∞xn∑k=1n1−limn→∞x2n2∑k=1nk+limn→∞x3n3∑k=1nk+limn→∞x4n4∑k=1nk3+…  ∑k=1nkm=nm+1m+1+amnm+…+a1n=nm+1m+1+∑j=1majnj  After solving limits. We get:  f(x)=x1−x22+x33−x44+…=∑k=1∞(−1)k+1xkk=ln⁡(x+1)  f(x)=ln⁡(x+1)  f(1)=ln⁡(2)

RizerMix · Accepted Answer

Note that we can write ∑k=1N1k+N=∑k=N+12N1k =∑k=12N1k−∑k=1N1k Next, we split the first sum on the right-hand side into sums of the even terms and odd terms and find that ∑k=12N1k−∑k=1N1k=(∑k=1N12k+∑k=1N12k−1)−∑k=1N1k =∑k=1N12k−1−∑k=1N12k =∑k=12N(−1)k−1k (1) Letting N→∞ in (1), we find limN→∞∑k=1N1k+N=limN→∞∑k=12N(−1)k−1k =∑k=1∞(−1)k−1k =log⁡(2) as was to be shown!

I have some difficulty to prove the following limit: \lim_{N\to\infty}\sum_{k=1}^N\frac{1}{k+N}=\ln(2)

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