"\\sum\\limits_{j=1}^n a_j"
Since
"a_j=a_{j-1}r=a_1r^{j-1}, r\\ne 1"
we assume that
"\\sum\\limits_{j=1}^n a_j=\\sum\\limits_{j=1}^n a_1r^{j-1}=a_1\\sum\\limits_{j=1}^n r^{j-1}"
It is known, that
"(1-r^n)=(1-r)(1+r+r^2+...+r^{n-1})"
Hence
"\\sum\\limits_{j=1}^n r^{j-1}=\\frac{1-r^n}{1-r}"
Therefore
"\\sum\\limits_{j=1}^n a_j=a_1\\frac{1-r^n}{1-r}"
To find the sum of the series we need to find
"\\lim\\limits_{n\\to \\infty}\\sum\\limits_{j=1}^n a_j=\\lim\\limits_{n\\to \\infty}a_1\\frac{1-r^n}{1-r}=\\left\\{\n\\begin{matrix}\n \\infty , & r>1 \\\\\n \\frac{a_1}{1-r}, & r<1\n\\end{matrix} \n\\right."
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