<s>[INST] Can anyone help to proof convergence and find the sum of such series? May be correct my mistakes. $$1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}-\frac{1}{10^{2}}-\frac{1}{10^{2}\sqrt{10}}+\cdot\cdot\cdot$$
I personally have such an idea: try to make geometric series like this
$$1-\frac{1}{\sqrt{10}}(1-\frac{1}{10}+\frac{1}{100}+\cdot\cdot\cdot)-\frac{1}{10}(1+\frac{1}{10}+\cdot\cdot\cdot)$$ using  3 5 7 terms and 2 4 6 terms to form two geometric series. then if we calculate sum for them we will have $$S_1 = \frac{10}{11}$$ and $$S_2 = \frac{10}{9}$$ and after substituting them we'll finally have sum of series $$S = \frac{88-9\sqrt{10}}{99}$$
Is everything right here?
But how to check it for convergence?
 [/INST] Write your sum like: 
$$S = {S}_1 + {S}_2 + {S}_3$$
Where, 
$${S}_1 = 1 + \frac{1}{\sqrt{10}\cdot 10} + \frac{1}{10^{3}} + ... = \frac{1}{1 - \frac{1}{10 \cdot\sqrt(10)}} = \frac{10\sqrt{10}}{10\sqrt{10} - 1}$$
$${S}_2 = -\frac{1}{\sqrt{10}} - \frac{1}{10^{2}} - ... = -\frac{1}{\sqrt{10}} \cdot{S}_1$$
$${S}_3 = -\frac{1}{10} - \frac{1}{10^{2}\sqrt{10}} - ... = -\frac{1}{10}\cdot{S}_1$$
And finally:
$$S = \frac{9\sqrt{10} - 10}{10\sqrt{10} - 1}$$
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