Could you explain how to find this limit?  $\lim_{n \rightarrow \infty} \sum^n_{k=1} \frac{1}{k+n}$
9339
What is the sum of the 'second half' of the harmonic series?  $$\lim_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} =~ ?$$  More precisely, what is the limit of the above sequence of partial sums?
29097
Calculate below limit $$\lim_{n\to\infty}\left(\sum_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$$
5154
What is the result of the next limit:   $$ \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}},\ k \in \mathbb{R} $$  Why (theorem)?
51385
It seems that I'm missing something about this.  First of all, the series is convergent: $\lim_{n\rightarrow\infty}\frac{2^{-n-1} (n+1)^2}{2^{-n} n^2}=\frac{1}{2}$ (ratio test)  What I tried to do is to find a limit of a partial sum $\lim_{n\rightarrow\infty}S_n$ as follows: $S_n=\frac{\frac{1}{6} n (n+1) (2 n+1)}{\frac{1-\left(\frac{1}{2}\right)^n}{2 \left(1-\frac{1}{2}\right)}}$. Still, the limit is $\infty$ and I'm clearly doing something wrong.
114120
How can I calculate this: $\lim\limits_{n \rightarrow \infty} \frac{1+\sqrt[n] {e}+\sqrt[n] {e^2}+ \dots + \sqrt[n] {e^{n-1}}  } {n}$? I how come so far: $$ \lim_{n \rightarrow \infty} \frac{1+\sqrt[n] {e}+\sqrt[n] {e^2}+ \dots + \sqrt[n] {e^{n-1}}  } {n} = \lim_{n \rightarrow \infty} \frac{1}{n} \left(1 +\sqrt[n] {e}+\sqrt[n] {e^2}+ \dots + \sqrt[n] {e^{n-1}}\right) = $$  $$ \lim_{n \rightarrow \infty}\frac{1}{n}\sum_{i =1}^{n}e^{\frac{i-1}{n}}$$ But how do I continue?
11141
Direct way leads to nowhere.  $$\lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {n^2 +i} } = \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{n \sqrt {1 +\frac{i}{n^2}} } = $$  $$ = \lim_{n\rightarrow \infty} \frac{1}{n} \cdot \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {1 +\frac{i}{n^2}} } = 0 \cdot \sum^{n}_{i=1} 1 = 0 \cdot n $$  So, I tried to apply squeeze theorem:  $$\lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {n^2 +i} } \leq \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{n}  = 1$$  Could you help me to figure out the lower bound?
79147
$$ \lim_{n \rightarrow \infty} \sum_{k=1}^n \left( \sqrt{n^4+k}\ \sin\frac{2k\pi}{n} \right) = ? $$  I tried to transform it into a Riemann sum, to use Taylor-series of the sine function, to estimate, but nothing. Any help would be great.
107026
$\lim_{x\to\infty} a_n$ defined as $a_{n+1}=a_n+\frac{1}{a_n}$ with $a_1=1$.  I've tried writing out the first few terms and this sum is increasing, but progressively less-so over time. I'm just not certain how to go about caclulating the limit here. I feel that it does have a finite limit, but I'm not certain of what it is.
104005
Let $x_0 = 1, x_1 = \frac{1}{1+x_0}, x_2 = \frac{1}{1+x_1} \cdots x_n = \frac{1}{1+x_{n-1}}$  Find $\lim_{n\rightarrow \infty} x_n$ as $n$ approaches infinity.  I don't know to represent this recurrence relation in a way that I can solve for $x_n$ and take the limit. Is there another way that I can do this?
53946
Prove: $\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^{n}\sqrt{1+\frac{k}{n}}=\frac{2}{3}(2\sqrt{2}-1)$  What method to use in order to find the closed form of summation $\sum\limits_{k=1}^{n}\sqrt{1+\frac{k}{n}}$?
7032
I've spent the better part of this day trying to show from first principles that this sequence tends to 1. Could anyone give me an idea of how I can approach this problem?  $$ \lim_{n \to +\infty} n^{\frac{1}{n}} $$
110152
I need to prove that $$\lim_{n\rightarrow \infty} \frac{n^k}{2^n}=0$$ where $k\in \mathbb{N}$. All I can think of is to use something like L'Hopital's rule but I suppose there must be a another simpler way. I would much appreciate if someone could give me a hint. Thanks
94914
I am trying to find $$\lim_{n\rightarrow \infty}\prod_{k=2}^{n}\frac{k^3-1}{k^3+1}.$$I have no idea how I can start. Please help me.  Thanks!
42353
I’m having a problem approaching the question. The fact that it’s a limit of a sum is confusing to me.   $$\lim _{n\to \infty }\left(\frac{1}{n^2+1}+\frac{1}{n^2+2}+...+\frac{1}{n^2+n}\right)$$
114099
Given the Series $$\sum_{k=1}^\infty \frac{1}{k(k+2)}$$  How exactly would I find out the limit is $\frac34$ as suggested by Wolframalpha? I already found out I can prove it actually converges by performing the comparison test and seeing that the underlying sequence isn't a null-sequence. But unfortunately I am absolutely clueless on how to prove that it converges to $\frac34$.  Regards,  Dennis
78161
I'm trying to find the limit of $a_n = \left(1-\frac{1}{n^2}\right)^n$ for $n \rightarrow \infty$.  It seems that the limit is $1$, since $a_n = 0.999...$ for large $n$. The presentation $a_n = \frac{(n^2-1)^n}{n^{2n}}$ and expanding was my first idea, but I couldn't get the result from there. Any ideas?
109362
Evaluate the limit      $$ \lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} $$   My Attempt:  $$ \lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} = \lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^{n^2}\cdot \frac{1}{e^n} $$  How can I solve the problem from this point?