How to write the first five terms of this series in the following form by MMA code?
$\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot(2 n-1)}{2 \cdot 4 \cdot \cdots \cdot 2 n}= \frac{1}{2}+\frac{1 \cdot 3}{2 \cdot 4}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{2 \cdot 4 \cdot 6 \cdot 8}+\frac{1 \cdot 3 \cdot 5 \cdot 7 \cdot 9}{2 \cdot 4 \cdot 6 \cdot 8 \cdot 10}$
I tried to write code with HoldForm, but I couldn't get the result.
For example,
$\sum_{n=1}^{\infty} \frac{1+n}{1+n^{2}}$
Sum[(HoldForm[(1 + #1)/(1 + #1^2)] & )[i], {i, 1, 5}] $\frac{1+1}{1+1^{2}}+\frac{1+2}{1+2^{2}}+\frac{1+3}{1+3^{2}}+\frac{1+4}{1+4^{2}}+\frac{1+5}{1+5^{2}}$
Or,
Total[Apply[HoldForm[#1/#2] & , Table[{1 + i, 1 + i^2}, {i, 1, 5}], {1}]] $\frac{2}{2}+\frac{3}{5}+\frac{4}{10}+\frac{5}{17}+\frac{6}{26}$
Update: Some other questions
How to get the results in the following form?
$\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot \cdots \cdot(2 n-1)}{2 \cdot 4 \cdot \cdots \cdot 2 n}=\frac{1}{2}+\frac{3}{8}+\frac{5}{16}+\frac{35}{128}+\frac{63}{256}$
$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{5^{n}}=\frac{1}{5}-\frac{1}{5^{2}}+\frac{1}{5^{3}}-\frac{1}{5^{4}}+\frac{1}{5^{5}}=\frac{1}{5}-\frac{1}{25}+\frac{1}{125}-\frac{1}{625}+\frac{1}{3125}$
$\sum_{n=1}^{\infty} \frac{1+n}{1+n^{2}}= 1+\frac{3}{5}+\frac{2}{5}+\frac{5}{17}+\frac{3}{13}$
