# how to integrate a sum series?

Posted 3 years ago
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 Hi, given that the “introduce yourself “ board is temporarly unavailable I briefly introduce myself here and then I submit my question about calculus and WA.My name is Eugenio and I work in Human resources sector in Italy. I'm also a part time forex and index trader. I'm interested in building a deep knowledge in quantitative finance and historical data analysis and handling. I'm currently rebuilding my math foundations and I'm using WA as a tool to achieve a deeper understanding of several math skills. I'll be grateful to all the people that will help me to become a WA power user.Here is my question: I have to calculate the definite integral from 0 to 1 of a series sum from n= 1 to infinity. The function is ((n+2)/(2^n))x^(n+1) I've tried to use this syntax in WA search engine:integrate [sum_n=0^infinity ((n+2)/(2^n))x^(n+1) {x, 0, 1} series representationI have written “series representation” at the end cause I need the output to be in series form. WA doesen't understand the query. Did I make some mistake? Is possible to use WA to make short scripts? What I mean is: an instruction to assign the series sum to a variable; successive instructions to perform on that variable different calculations.tnx Answer
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Posted 3 years ago
 Here is my question: I have to calculate the definite integral from 0 to 1 of a series sum from n= 1 to infinity. The function is ((n+2)/(2^n))x^(n+1) I've tried to use this syntax in WA search engine: I am not too familiar with the Wolfram Alpha language syntax. I only use the Wolfram Mathematica language, and the syntax in that language is f = ((n + 2)/(2^n)) x^(n + 1); fi = Integrate[f, {x, 0, 1}]; Sum[fi, {n, 1, Infinity}] And the answer is 1. May be the above will help you translate that to the Wolfram Alpha syntax more easily now. Wolfram Alpha should also accept Wolfram Mathematica syntax as far as I know. So the above should works as is inside Wolfram Alpha but I have not tried it. Answer
Posted 3 years ago
 Thank You Nasser, however WA only find the roots of the equation and than says to me that it has exceeded computational time. It seems to me that WA is only an appetizer towards Mathematica which I think is indeed an interesting software to buy. Answer
Posted 3 years ago
 however WA only find the roots of the equation I am not sure if you typed the correct commands, because there should not be any root finding there. It is just one Integrate command and one Sum command.If you are interested in Wolfram Language, I would recommend getting Mathematica itself. They also have trial version that you can download and try for free. I think it is 15 days free trial. Answer
Posted 3 years ago
 Hi,alternatively, you can use a free tier of the Development Cloud. This gives: Cheers,Marco Answer
Posted 3 years ago
 Hi Marco, tnx for Your suggestion, it works. However this is not the only problem. I needed the output in this form: 1/(2^0)+1/(2^1)+1/(2/2)+...+1/(2^n) Is there some costruct either in Wa or in Mathematica that gives me the answer in that form? Tnx :-) Eugenio Answer