Question Number 125006 by Mammadli last updated on 07/Dec/20 | ||
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$$\boldsymbol{{prove}}\:\boldsymbol{{that}}: \\ $$ $$\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+...+\frac{\mathrm{1}}{\mathrm{127}}+\frac{\mathrm{1}}{\mathrm{128}}>\mathrm{1} \\ $$ | ||
Commented bymr W last updated on 07/Dec/20 | ||
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$${that}'{s}\:{too}\:{easy}. \\ $$ $${try}\:{to}\:{prove} \\ $$ $$\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+...+\frac{\mathrm{1}}{\mathrm{127}}+\frac{\mathrm{1}}{\mathrm{128}}>\mathrm{3} \\ $$ | ||
Commented byMammadli last updated on 07/Dec/20 | ||
Sorry dear, >3 | ||