# Tag Info

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### Limit of $\sum_{k=1}^n \frac{1}{k+1/2} - \ln(n+1/2)$

Considering the hint provided by @openspace, I have got the correct result. \begin{align*} \lim_{n\to\infty}\sum_{k=1}^n \frac{1}{k+\frac{1}{2}} - \ln(n+\frac{1}{2}) &= \lim_{n\to\...
• 105
Accepted

### Limit of $\sum_{k=1}^n \frac{1}{k+1/2} - \ln(n+1/2)$

Let $\psi(z)=\Gamma'(z)/\Gamma(z)$. Then it is known that $$\psi(z+1)=-\gamma+\sum_{n=1}\left(\frac1n-{1\over n+z}\right),$$ so there is \begin{aligned} \lim_{n\to+\infty}\left(\sum_{k=1}^n{1\over k+...
• 7,193
1 vote
Accepted

### Help understanding Spivak's solution, and a verification of my proof. Spivak Chapter 5, Question 3(vi)

OP's analysis is fine, but he makes a small mistake. On the one hand we have $1-\epsilon < \sqrt x < 1 + \epsilon$, which after squaring leads to $(1-\epsilon)^{2} < x < (1+\epsilon)^{2}$. ...
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### Computing the Limit $\lim_{n \to \infty} n\left[ \frac{a_{n+1}}{a_{n}} - \left(\frac{n}{n+1}\right)^{\frac{1}{3}} \right]$

The expression can be represented as $${n\over 2(n+1)}-\left ({n\over n+1}\right )^{4/3}\,[-(n+1)]\left [\left (1-{1\over n+1}\right )^{2/3}-1\right ]$$ The limit of the last two factors is equal by ...
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### Computing the Limit $\lim_{n \to \infty} n\left[ \frac{a_{n+1}}{a_{n}} - \left(\frac{n}{n+1}\right)^{\frac{1}{3}} \right]$

Let me call your while expression $A_n$. Notice, that there is the identity $\Gamma(1/2 + n) = \sqrt{\pi} \frac{(2n)!}{n! 4^n}$. Then you have $$a_n = \frac{\Gamma(n+1/2)}{\sqrt{\pi} n!}$$ and yields ...
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### Showing that the function that returns $1$ on the rationals and $0$ elsewhere has no limit at $0$. Is there anything wrong in this proof?

Added a bit of annotations to the proposed proof attempting to point out some logical flaws. Proof: Suppose that $\lim_{x \to 0; x\in \mathbb R}f(x) = L$. Then for every $\varepsilon > 0$, there ...
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