query: How to evaluate the sum $\sum_{i=1}^n{{n}\choose{i}}\frac{1}{i}$
positive_passages.docid: 35381
positive_passages.text: Sum with binomial coefficients: $\sum_{k=1}^m  \frac{1}{k}{m \choose k} $
negative_passages.docid: 84147
negative_passages.text: Evaluating $\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}$
negative_passages.docid: 28378
negative_passages.text: Find the infinite sum of the series $\sum_{n=1}^\infty \frac{1}{n^2 +1}$
negative_passages.docid: 3341
negative_passages.text: Find $\sum_{r=1}^{3n-1}{ (-1)^{r-1}r\over{3n \choose r}}$, if $n$ is even
negative_passages.docid: 43903
negative_passages.text: Find the sum for $\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{2n+1}{3^n} $
negative_passages.docid: 72502
negative_passages.text: Evaluate $\sum_{n=1}^\infty nx^{n-1}$
negative_passages.docid: 39909
negative_passages.text: Getting a closed form from $\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1$
negative_passages.docid: 121554
negative_passages.text: Finding $\sum\frac{1}{2^n(n+1)}$
negative_passages.docid: 47706
negative_passages.text: Find the value of $ \sum_{n=1}^{\infty} \frac{n^3}{3^n} $
negative_passages.docid: 86426
negative_passages.text: calculate $\sum_{1\le i\le r}\frac{i+1} { r+1}{2r-i\choose r-i}{s+i-2\choose i}+\frac{1}{r+1}{2r\choose r}$?
negative_passages.docid: 6838
negative_passages.text: Evaluate the series $S_n = \sum_{k=1}^n\log\frac {k (k + 2)}{(k + 1)^2}$
negative_passages.docid: 16967
negative_passages.text: Sum : $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}$
negative_passages.docid: 91784
negative_passages.text: Calculating $\sum_{n=1}^\infty {\frac{1}{2^nn(3n-1)}}$
negative_passages.docid: 138542
negative_passages.text: Evaluating sums question $\sum_{k=2}^n \frac{1}{k^2-1}$
negative_passages.docid: 106028
negative_passages.text: Sum of $\frac{1}{(2n)(2n+1)}$ from $1$ to infinity
negative_passages.docid: 115420
negative_passages.text: Evaluate $\sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}$
negative_passages.docid: 20317
negative_passages.text: Proving $\sum_{k=1}^{n}{(-1)^{k+1} {{n}\choose{k}}\frac{1}{k}=H_n}$