Indian mathematician (1887–1920)
Srinivasa Aiyangar Ramanujan (Tamil: ஸ்ரீனிவாஸ ஐயங்கார் ராமானுஜன்) (22 December 1887 – 26 April 1920) was an Indian mathematician and autodidact, noted for his extraordinary achievements in the field of mathematical analysis, number theory, infinite series, and continued fractions. In his uniquely self-developed mathematical research he not only rediscovered known theorems but also produced brilliant new work, prompting his mentor G. H. Hardy to compare his brilliance to that of Euler and Gauss. He became a Fellow of the Royal Society, and India now observes his birthday as National Mathematics Day.
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I beg to introduce myself to you as a clerk in the Accounts Department of the Port Trust Office at Madras... I have no University education but I have undergone the ordinary school course. After leaving school I have been employing the spare time at my disposal to work at Mathematics. I have not trodden through the conventional regular course which is followed in a University course, but I am striking out a new path for myself. I have made a special investigation of divergent series in general and the results I get are termed by the local mathematicians as "startling". ...Very recently I came across a tract published by you styled Orders of Infinity in page 36 of which I find a statement that no definite expression has been as yet found for the number of prime numbers less than any given number. I have found an expression which very nearly approximates to the real result, the error being negligible. I would request that you go through the enclosed papers. Being poor, if you are convinced that there is anything of value I would like to have my theorems published. I have not given the actual investigations nor the expressons that I get but I have indicated the lines on which I proceed. Being inexperienced I would very highly value any advice you give me. Requesting to be excused for the trouble I give you. I remain, Dear Sir, Yours truly...
If <math>n</math> is any positive quantity shew that
<math>\frac 1{n} > \frac 1{n+1} + \frac 1{{(n+2)}^2} + \frac 3{{(n+3)}^3} + \frac {4^2}{{(n+4)}^4} + \frac {5^3}{{(n+5)}^5} + \dots</math> Find the difference
approximately when <math>n</math> is great.
Hence shew that
<math>\frac 1{1001} + \frac 1{1002^2} + \frac 3{1003^3} + \frac {4^2}{1004^4} + \frac {5^3}{1005^5} + \dots < \frac 1{1000}</math> by <math>10^{-440}</math> nearly.